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All questions of Mathematical Induction for JEE Exam
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
a)
b)
c)
d)
|
Raghav Bansal answered |
A = {(a,1) (0,a)}
A2 = {(a,1) (0,a)} * {(a,1) (0,a)}
= {(a2+0, a+a) (0+0, a2)}
A2 = {(a2, 2a) (0, a2)}
A * A2 = A3
A3 = {(a2, 2a) (0, a2)} * {(a,1) (0,a)}
= {(a3+0, a2+2a2) (0+0, 0+a3)}
= {(a3, 3a2) (0, a3)}
Therefore An => {(an, 3a(n-1)) (0, an)}
A2 = {(a,1) (0,a)} * {(a,1) (0,a)}
= {(a2+0, a+a) (0+0, a2)}
A2 = {(a2, 2a) (0, a2)}
A * A2 = A3
A3 = {(a2, 2a) (0, a2)} * {(a,1) (0,a)}
= {(a3+0, a2+2a2) (0+0, 0+a3)}
= {(a3, 3a2) (0, a3)}
Therefore An => {(an, 3a(n-1)) (0, an)}
The greatest positive integer, which divides (n+1)(n+2)(n+3)..................(n+r) ∀n∈W, is- a)(r + 1) !
- b)r
- c)n+r
- d)r !
Correct answer is option 'D'. Can you explain this answer?
The greatest positive integer, which divides (n+1)(n+2)(n+3)..................(n+r) ∀n∈W, is
a)
(r + 1) !
b)
r
c)
n+r
d)
r !
|
Defence Exams answered |
P(n) = (n+1)(n+2).........(n+r)
Now find p(1)
P(1) = r! (r+1)
Similarly
P(k) will be divisible by r!
P(k+1) is also divisible by r!
Hence p(n) will also be divisible by r!.
Proving P(1) to be true is known as _______ .- a)Basic Step
- b)Inductive Step
- c)Hypothesis step
- d)All of the above
Correct answer is option 'A'. Can you explain this answer?
Proving P(1) to be true is known as _______ .
a)
Basic Step
b)
Inductive Step
c)
Hypothesis step
d)
All of the above
|
|
Upasana Sengupta answered |
Explanation:
The process of mathematical induction is used to prove that a statement is true for all positive integers. It has three steps: the basic step, the inductive step, and the conclusion.
Basic Step:
The basic step is the first step in mathematical induction. It involves proving that the statement is true for the first positive integer, usually 1. This step is also known as the base case. If the statement is not true for the base case, then the proof fails.
Inductive Step:
The inductive step is the second step in mathematical induction. It involves assuming that the statement is true for some arbitrary positive integer k, and then proving that the statement is also true for the next positive integer, k+1. This step is also known as the inductive hypothesis.
Conclusion:
The conclusion is the final step in mathematical induction. It involves using the basic step and the inductive step to conclude that the statement is true for all positive integers. This step is also known as the inductive conclusion.
Therefore, proving P(1) to be true is known as the Basic Step of mathematical induction.
The process of mathematical induction is used to prove that a statement is true for all positive integers. It has three steps: the basic step, the inductive step, and the conclusion.
Basic Step:
The basic step is the first step in mathematical induction. It involves proving that the statement is true for the first positive integer, usually 1. This step is also known as the base case. If the statement is not true for the base case, then the proof fails.
Inductive Step:
The inductive step is the second step in mathematical induction. It involves assuming that the statement is true for some arbitrary positive integer k, and then proving that the statement is also true for the next positive integer, k+1. This step is also known as the inductive hypothesis.
Conclusion:
The conclusion is the final step in mathematical induction. It involves using the basic step and the inductive step to conclude that the statement is true for all positive integers. This step is also known as the inductive conclusion.
Therefore, proving P(1) to be true is known as the Basic Step of mathematical induction.
If P(n) : (2n + 7) < (n + 3)2 then P(3) is- a)(2 x 3 + 7 x 3) < (3 + 3)2
- b)(23 + 7) < (n + 3)2
- c)(2 x 3 + 7) < (3 + 3)2
- d)(32 + 7) < (2 + 7)2
Correct answer is option 'C'. Can you explain this answer?
If P(n) : (2n + 7) < (n + 3)2 then P(3) is
a)
(2 x 3 + 7 x 3) < (3 + 3)2
b)
(23 + 7) < (n + 3)2
c)
(2 x 3 + 7) < (3 + 3)2
d)
(32 + 7) < (2 + 7)2
|
Suresh Iyer answered |
P(n) : (2n + 7) < (n + 3)2
P(3) = [(2(3) + 7) < (3+ 3)2
P(3) = [(2(3) + 7) < (3+ 3)2
For each n ∈ N, 72n + 16n – 1 is divisible by :- a)34
- b)54
- c)66
- d)64
Correct answer is option 'D'. Can you explain this answer?
For each n ∈ N, 72n + 16n – 1 is divisible by :
a)
34
b)
54
c)
66
d)
64
|
|
Suresh Iyer answered |
For n=1, we have 72 +16 -1 = 64 which is divisible by 64
For n=k, we assume that 72k + 16k-1 is divisible by 64
We must check for n=k+1, is the statement true
72(k+1) +16(k+1) -1 = 49 * 72k + 16k +15
we adjust this expression by adding and subtracting terms, that is
we add 49*16k, -49 but we have to subtract 49*16k and -49
we have the following
49*72k +16k +15 = 49*72k + 49*16k -49 -49*16k -49 +16k +15 =
49(72k +16k -1) - 49(16k-1) +16k +15 =
49(7k +16k -1) - 48*16k +64
all these terms are divisible by 64
For n=k, we assume that 72k + 16k-1 is divisible by 64
We must check for n=k+1, is the statement true
72(k+1) +16(k+1) -1 = 49 * 72k + 16k +15
we adjust this expression by adding and subtracting terms, that is
we add 49*16k, -49 but we have to subtract 49*16k and -49
we have the following
49*72k +16k +15 = 49*72k + 49*16k -49 -49*16k -49 +16k +15 =
49(72k +16k -1) - 49(16k-1) +16k +15 =
49(7k +16k -1) - 48*16k +64
all these terms are divisible by 64
then A10 equals
For each n ∈ N , n (n + 1) (2n + 1) is divisible by :- a)2
- b)8
- c)6
- d)15
Correct answer is option 'C'. Can you explain this answer?
For each n ∈ N , n (n + 1) (2n + 1) is divisible by :
a)
2
b)
8
c)
6
d)
15
|
Anuj Datta answered |
Explanation:
To prove that n(n+1)(2n+1) is divisible by 6, we need to show that it is divisible by 2 and 3.
Divisibility by 2:
- A number is divisible by 2 if it is even.
- If n is even, then n(n+1)(2n+1) is certainly even.
- If n is odd, then (n+1) is even, and (2n+1) is odd.
- The product of an even number and an odd number is always even.
- Therefore, n(n+1)(2n+1) is divisible by 2.
Divisibility by 3:
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- Let's consider the three factors separately:
- n: If n is divisible by 3, then its sum of digits is divisible by 3.
- (n+1): If n is not divisible by 3, then (n+1) is divisible by 3, as it is the next consecutive number.
- (2n+1): If n is not divisible by 3, then (2n+1) is divisible by 3, as it is the next odd number.
- Therefore, one of the three factors, n, (n+1), or (2n+1), is divisible by 3.
- So, n(n+1)(2n+1) is divisible by 3.
Combining the Divisibility:
- We have shown that n(n+1)(2n+1) is divisible by both 2 and 3.
- Any number that is divisible by both 2 and 3 is divisible by their product, i.e., 6.
- Therefore, n(n+1)(2n+1) is divisible by 6.
Conclusion:
- From the above explanation, it can be concluded that n(n+1)(2n+1) is divisible by 6.
- Thus, the correct answer is option 'C'.
To prove that n(n+1)(2n+1) is divisible by 6, we need to show that it is divisible by 2 and 3.
Divisibility by 2:
- A number is divisible by 2 if it is even.
- If n is even, then n(n+1)(2n+1) is certainly even.
- If n is odd, then (n+1) is even, and (2n+1) is odd.
- The product of an even number and an odd number is always even.
- Therefore, n(n+1)(2n+1) is divisible by 2.
Divisibility by 3:
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- Let's consider the three factors separately:
- n: If n is divisible by 3, then its sum of digits is divisible by 3.
- (n+1): If n is not divisible by 3, then (n+1) is divisible by 3, as it is the next consecutive number.
- (2n+1): If n is not divisible by 3, then (2n+1) is divisible by 3, as it is the next odd number.
- Therefore, one of the three factors, n, (n+1), or (2n+1), is divisible by 3.
- So, n(n+1)(2n+1) is divisible by 3.
Combining the Divisibility:
- We have shown that n(n+1)(2n+1) is divisible by both 2 and 3.
- Any number that is divisible by both 2 and 3 is divisible by their product, i.e., 6.
- Therefore, n(n+1)(2n+1) is divisible by 6.
Conclusion:
- From the above explanation, it can be concluded that n(n+1)(2n+1) is divisible by 6.
- Thus, the correct answer is option 'C'.
If p(n) is the statement : 12 + 32 + 52 + __ __ + (2n - 1)2 
then which one of the following is incorrect?- a)P(3) is true
- b)P(1) is true
- c)P(4) is true
- d)P(5) is true
Correct answer is option 'B'. Can you explain this answer?
If p(n) is the statement : 12 + 32 + 52 + __ __ + (2n - 1)2 then which one of the following is incorrect?
then which one of the following is incorrect?a)
P(3) is true
b)
P(1) is true
c)
P(4) is true
d)
P(5) is true
|
Chirag Verma answered |
Correct Answer :- b
Explanation : Let P(n) be the given statement, i.e., P(n) : (2n + 1) < 2n
for all natural
numbers, n ≥ 3. We observe that P(3) is true, since
2.3 + 1 = 7 < 8 = 23
Assume that P(n) is true for some natural number k, i.e., 2k + 1 < 2k
To prove P(k + 1) is true, we have to show that 2(k + 1) + 1 < 2k+1. Now, we have
2(k + 1) + 1 = 2 k + 3
= 2k + 1 + 2 < 2k + 2 < 2k
. 2 = 2k + 1
Thus P(k + 1) is true, whenever P(k) is true.
Hence, by the Principle of Mathematical Induction P(n) is true for all natural numbers, n ≥ 3
Let P(n) is the statement the product of two consecutive natural numbers is even. Then which of the following is true?- a)P(100)
- b)P(n), n ∈ N
- c)P(20)
- d)P(3)
Correct answer is option 'B'. Can you explain this answer?
Let P(n) is the statement the product of two consecutive natural numbers is even. Then which of the following is true?
a)
P(100)
b)
P(n), n ∈ N
c)
P(20)
d)
P(3)
|
Ram Prakash answered |
The product of natural number is natural.and all even number be natural number
The principle of mathematical induction is for the set of:- a)Rational numbers
- b)Positive integers
- c)Whole number
- d)Integers
Correct answer is 'B'. Can you explain this answer?
The principle of mathematical induction is for the set of:
a)
Rational numbers
b)
Positive integers
c)
Whole number
d)
Integers
|
|
Riya Banerjee answered |
A class of integers is called hereditary if, whenever any integer x belongs to the class, the successor of x (that is, the integer x + 1) also belongs to the class. The principle of mathematical induction is then: If the integer 0 belongs to the class F and F is hereditary, every nonnegative integer belongs to F.
2n > n2 when n ∈ N such that- a)n > 2
- b)n < 5
- c)n > 3
- d)n ≥ 5
Correct answer is option 'D'. Can you explain this answer?
2n > n2 when n ∈ N such that
a)
n > 2
b)
n < 5
c)
n > 3
d)
n ≥ 5
|
Lavanya Menon answered |
case: n=5 (the problem states "n greater than 4", so let's pick the first integer that matches)
25 > 52
⟹32 > 25 - ok!
Now, Inductive Step:
2n+1 > (n+1)2
now expanding
2∗2n > n2 + 2n + 1
= (n+1)2
The first inequality follows from the induction hypothesis and as for the second, we know that (n−1)2 ≥ 42 > 2, since n ≥ 5. We can expand this inequality (n−1)2 > 2 as follows:
n2−2n+1 > 2
n2−2n−1 > 0
2n2−2n−1 > n2
2n2 > n2+2n+1
= (n+1)2,
25 > 52
⟹32 > 25 - ok!
Now, Inductive Step:
2n+1 > (n+1)2
now expanding
2∗2n > n2 + 2n + 1
= (n+1)2
The first inequality follows from the induction hypothesis and as for the second, we know that (n−1)2 ≥ 42 > 2, since n ≥ 5. We can expand this inequality (n−1)2 > 2 as follows:
n2−2n+1 > 2
n2−2n−1 > 0
2n2−2n−1 > n2
2n2 > n2+2n+1
= (n+1)2,
The sum of cubes of three consecutive natural numbers is divisible by:- a)2
- b)9
- c)7
- d)5
Correct answer is option 'A'. Can you explain this answer?
The sum of cubes of three consecutive natural numbers is divisible by:
a)
2
b)
9
c)
7
d)
5
|
Gaurav Kumar answered |
n3 + (n+1)3 + (n−1)3
=3(n)3 + 6n
=3n(n2 + 2)
=3n(n2 − 1 + 3)
=3(n(n−1)(n+1)+3n)
n(n−1)(n+1) and 3n are both divisible by 3. Hence their sum is also divisible by 3.
Therefore the whole number is divisible by 9.
=3(n)3 + 6n
=3n(n2 + 2)
=3n(n2 − 1 + 3)
=3(n(n−1)(n+1)+3n)
n(n−1)(n+1) and 3n are both divisible by 3. Hence their sum is also divisible by 3.
Therefore the whole number is divisible by 9.
The smallest positive integer n, for which 
- a)2
- b)3
- c)4
- d)1
Correct answer is option 'A'. Can you explain this answer?
The smallest positive integer n, for which
a)
2
b)
3
c)
4
d)
1
|
Pioneer Academy answered |
n! < [(n+1)/2]n
For(n=1) 1! < [(1+1)/2]1
= 1 < (2/2)1
1 < 1 (False)
For(n=2) 2! < [(2+1)/2]2
= 2 < (3/2)2
1 < 6/4
1 < 3/2
1 < 1.5 (True)
Therefore smallest number will be 2.
For(n=1) 1! < [(1+1)/2]1
= 1 < (2/2)1
1 < 1 (False)
For(n=2) 2! < [(2+1)/2]2
= 2 < (3/2)2
1 < 6/4
1 < 3/2
1 < 1.5 (True)
Therefore smallest number will be 2.
2.42n + 1 + 33n + 1 for all n is divisible by- a)3
- b)11
- c)2
- d)4
Correct answer is option 'B'. Can you explain this answer?
2.42n + 1 + 33n + 1 for all n is divisible by
a)
3
b)
11
c)
2
d)
4
|
Naina Bansal answered |
So, It is divisible by 11.
The greatest positive integer, which divides n (n + 1) (n + 2) (n + 3) for all n ∈ N, is- a)24
- b)120
- c)6
- d)2
Correct answer is option 'A'. Can you explain this answer?
The greatest positive integer, which divides n (n + 1) (n + 2) (n + 3) for all n ∈ N, is
a)
24
b)
120
c)
6
d)
2
|
|
Mysterio Man answered |
Let us make an equation -->
p(n):n(n+1)(n+2)(n+3)=24 n,where n belongs to natural number,
by putting the value of n=1,
LHS=1(2)(3)(4)=24=RHS,
so, p(n) is true for n=1,
assume that p(k) is true ,
p(k)= k(k+1)(k+2)(k+3)=24m--(1).,
we have to prove p(k+1) is true,
lhs for p(k+1)=,
(k+1)(k+2)(k+3) (k+4),
from 1,
LHS=24k where K is a natural number=RHS
p(n):n(n+1)(n+2)(n+3)=24 n,where n belongs to natural number,
by putting the value of n=1,
LHS=1(2)(3)(4)=24=RHS,
so, p(n) is true for n=1,
assume that p(k) is true ,
p(k)= k(k+1)(k+2)(k+3)=24m--(1).,
we have to prove p(k+1) is true,
lhs for p(k+1)=,
(k+1)(k+2)(k+3) (k+4),
from 1,
LHS=24k where K is a natural number=RHS
If n is a +ve integer, then 23n−7n−1 is divisible by- a)64
- b)36
- c)49
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If n is a +ve integer, then 23n−7n−1 is divisible by
a)
64
b)
36
c)
49
d)
none of these
|
|
Gaurav Kumar answered |
Which is divided by 49 by the case of above.
Assuming the truth of P(k) and proving P(k + 1) to be true, for some integer k is known as the _______ .- a)Basic Step
- b)nductive Step
- c)Hypothesis step
- d)All of the above
Correct answer is option 'B'. Can you explain this answer?
Assuming the truth of P(k) and proving P(k + 1) to be true, for some integer k is known as the _______ .
a)
Basic Step
b)
nductive Step
c)
Hypothesis step
d)
All of the above
|
Pooja Mehta answered |
The hypothesis in the inductive step, that the statement holds for some n, is called the induction hypothesis or inductive hypothesis. To prove the inductive step, one assumes the induction hypothesis and then uses this assumption, involving n, to prove the statement for n + 1.
In the principle of mathematical induction, which of the following steps is mandatory?- a)induction hypothesis
- b)inductive reference
- c)induction set assumption
- d)minimal set representation
Correct answer is option 'A'. Can you explain this answer?
In the principle of mathematical induction, which of the following steps is mandatory?
a)
induction hypothesis
b)
inductive reference
c)
induction set assumption
d)
minimal set representation
|
Ananya Das answered |
The hypothesis of Step is a must for mathematical induction that is the statement is true for n = k, where n and k are any natural numbers, which is also called induction assumption or induction hypothesis.
Every even power of an odd number greater than 1 when divided by 8 leaves 1 as the remainder.The inductive step for the above statement is- a)P(1) = 32 = 8.1 + 1
- b)P(k) = (2k + 1)2 = 8a + 1 implies P(k + 1):(2k + 3)2 = 8b + 1 for some natural numbers a,b
- c)P(k) = (2k + 1)2n = 8a + 1 implies P(k + 1) : (2k + 3)2n + 2 = 8b + 1 for some natural number a,b
- d)P(k) = (2k – 1)2 = 8a + 1 implies P(k + 1); (2k + 1)2 = 8b + 1 for some natural numbers a,b
Correct answer is option 'C'. Can you explain this answer?
Every even power of an odd number greater than 1 when divided by 8 leaves 1 as the remainder.The inductive step for the above statement is
a)
P(1) = 32 = 8.1 + 1
b)
P(k) = (2k + 1)2 = 8a + 1 implies P(k + 1):(2k + 3)2 = 8b + 1 for some natural numbers a,b
c)
P(k) = (2k + 1)2n = 8a + 1 implies P(k + 1) : (2k + 3)2n + 2 = 8b + 1 for some natural number a,b
d)
P(k) = (2k – 1)2 = 8a + 1 implies P(k + 1); (2k + 1)2 = 8b + 1 for some natural numbers a,b
|
|
Suresh Iyer answered |
If P(k) is the statement 23k – 1 is divisible by 7, then P(k + 1) is- a)23k – 2 is divisible by 7
- b)23k + 2 – 1 is divisible by 7
- c)23k + 1 – 1 is divisible by 7
- d)23k + 3 – 1 is divisible by 7
Correct answer is option 'C'. Can you explain this answer?
If P(k) is the statement 23k – 1 is divisible by 7, then P(k + 1) is
a)
23k – 2 is divisible by 7
b)
23k + 2 – 1 is divisible by 7
c)
23k + 1 – 1 is divisible by 7
d)
23k + 3 – 1 is divisible by 7
|
|
Saanvi Choudhury answered |
Let P(n): 23n - 1 is divisible by 7
Now, P( 1): 23 - 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 23k – 1 is divisible by 7.
or 23k -1 = 7m, m∈ N (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 23(k+1)– 1
= 23k.23 -1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.
Now, P( 1): 23 - 1 = 7, which is divisible by 7.
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
P(k): 23k – 1 is divisible by 7.
or 23k -1 = 7m, m∈ N (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 23(k+1)– 1
= 23k.23 -1
= 8(7 m + 1) – 1
= 56 m + 7
= 7(8m + 1), which is divisible by 7.
Thus, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for all natural numbers n.
If P(n) is the statement; 4 + 8 + 12 +……+ 4n = 2n (n + 1) then P(6) will be- a)4+8+.....+4(6) = 2(6)(6+1)
- b)4 + 8 + 12 …… + 4 x 4 = 2 x 4(4 + 1)
- c)4 + 8 + ……+ 6 = 2 x 6(6 + 1)
- d)none
Correct answer is option 'A'. Can you explain this answer?
If P(n) is the statement; 4 + 8 + 12 +……+ 4n = 2n (n + 1) then P(6) will be
a)
4+8+.....+4(6) = 2(6)(6+1)
b)
4 + 8 + 12 …… + 4 x 4 = 2 x 4(4 + 1)
c)
4 + 8 + ……+ 6 = 2 x 6(6 + 1)
d)
none
|
Tanvi Menon answered |
P(n) = 4 + 8 + 12 +……+ 4n = 2n (n + 1)
P(6) = 4 + 8 + 12 +……+ 4(6)
= 2(6)[(6) + 1)]
P(6) = 4 + 8 + 12 +……+ 4(6)
= 2(6)[(6) + 1)]
If n is a +ve integer, then 33n−26n−1 is divisible by- a)676
- b)627
- c)547
- d)239
Correct answer is option 'A'. Can you explain this answer?
If n is a +ve integer, then 33n−26n−1 is divisible by
a)
676
b)
627
c)
547
d)
239
|
|
Neha Joshi answered |
For n=1, 33n − 26n − 1
= 33 − 27 = 0
For n=2, 33n− 26n − 1
= 36 − 52 − 1
= 729 − 53
= 676
= 33 − 27 = 0
For n=2, 33n− 26n − 1
= 36 − 52 − 1
= 729 − 53
= 676
If n is a +ve integer, then 10n +3.4n+2 + 5 is divisible by- a)15
- b)4
- c)3
- d)9
Correct answer is option 'D'. Can you explain this answer?
If n is a +ve integer, then 10n +3.4n+2 + 5 is divisible by
a)
15
b)
4
c)
3
d)
9
|
|
Tanishq Ghoshal answered |
Explanation:
- If n is an even integer, then 10n is divisible by 4 and 4n is divisible by 8. So, 3.4n is divisible by 8.
- If n is an odd integer, then 10n is divisible by 5 and 4n is divisible by 4. So, 3.4n is divisible by 3.
- If n is divisible by 3, then 10n is divisible by 1000 and 4n is divisible by 27. So, 3.4n is divisible by 9.
- Therefore, 10n + 3.4n + 2 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore, 10n + 3.4n + 2 - 5 is divisible by 9 and 5. So, 10n + 3.4n + 2 - 5 is divisible by 9 and 5.
- Therefore,
If a statement is to be proved by mathematical induction, then the different steps necessary to prove it are- a)Prove Basic step and Inductive step
- b)Inductive step to be proved
- c)Prove P(1), P(2), P(3)
- d)Basic step to be proved
Correct answer is option 'A'. Can you explain this answer?
If a statement is to be proved by mathematical induction, then the different steps necessary to prove it are
a)
Prove Basic step and Inductive step
b)
Inductive step to be proved
c)
Prove P(1), P(2), P(3)
d)
Basic step to be proved
|
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Hardik Handa answered |
To prove a statement by induction, we must prove parts 1) and 2) above. The hypothesis of Step 1) -- "The statement is true for n = k" -- is called the induction assumption, or the induction hypothesis. It is what we assume when we prove a theorem by induction.
Let P(n) be a statement and let P(n)⇒P(n+1) for all natural numbers n, then P(n) is true for- a)all n > m , m being a fixed positive integer
- b)all n > 1
- c)nothing can be said
- d)all n
Correct answer is option 'D'. Can you explain this answer?
Let P(n) be a statement and let P(n)⇒P(n+1) for all natural numbers n, then P(n) is true for
a)
all n > m , m being a fixed positive integer
b)
all n > 1
c)
nothing can be said
d)
all n
|
Dipika Chauhan answered |
Given, P(n) = P(n+1)∀n∈N
Substituting n−1 in place of n,
P(n−1) = P(n)
Thus if P(k) is true for some k ∈ N, then it is true for k−1 and k+1.
Thus, it is true ∀ k ∈ N
Substituting n−1 in place of n,
P(n−1) = P(n)
Thus if P(k) is true for some k ∈ N, then it is true for k−1 and k+1.
Thus, it is true ∀ k ∈ N
The statement P (n) : “1 X 1! + 2 X 2! + 3 X 3! + …..+ n X n! = (n + 1) !..... 1 “ is- a)true for all n > 1
- b)for all n ∈ N
- c)not true for any n
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The statement P (n) : “1 X 1! + 2 X 2! + 3 X 3! + …..+ n X n! = (n + 1) !..... 1 “ is
a)
true for all n > 1
b)
for all n ∈ N
c)
not true for any n
d)
none of these
|
Sonal Kaur answered |
The statement P (n) does not have a clear definition or context. Can you please provide more information or clarify the statement?
If P (n) = 2+4+6+………………..+2n , n ∈ N , then P (k) = k (k + 1) + 2 ⇒ P (k + 1) = (k + 1) (k +2) + 2 for all k ∈ N. So we can conclude that P (n) = n (n + 1) +2 for- a)n > 2
- b)all n ∈ N
- c)n > 1
- d)nothing can be said
Correct answer is option 'D'. Can you explain this answer?
If P (n) = 2+4+6+………………..+2n , n ∈ N , then P (k) = k (k + 1) + 2 ⇒ P (k + 1) = (k + 1) (k +2) + 2 for all k ∈ N. So we can conclude that P (n) = n (n + 1) +2 for
a)
n > 2
b)
all n ∈ N
c)
n > 1
d)
nothing can be said
|
|
Neha R. P answered |
We note that P(1)=2 and hence,
P(n)=n(n+1)+2 is not true for n=1.
So, the principle of mathematical induction is not applicable and nothing can be said about the validity of the statement P(n)=n(n+1)+2.
Hence, the option D is correct.
P(n)=n(n+1)+2 is not true for n=1.
So, the principle of mathematical induction is not applicable and nothing can be said about the validity of the statement P(n)=n(n+1)+2.
Hence, the option D is correct.
If a,b,c ∈ N, an+bn is divisible by c, when n is odd but not when n is even, then the value of c is :- a)a - b
- b)a+b
- c)a3+b3
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If a,b,c ∈ N, an+bn is divisible by c, when n is odd but not when n is even, then the value of c is :
a)
a - b
b)
a+b
c)
a3+b3
d)
none of these
|
|
Disha Bajaj answered |
I'm sorry, but your question seems to be incomplete. Could you please provide more information or clarify your question?
The statement 3n>4n is true for all- a)n > 1 , n ∈ N
- b)n ∈ N
- c)n > 2 , n ∈ N
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The statement 3n>4n is true for all
a)
n > 1 , n ∈ N
b)
n ∈ N
c)
n > 2 , n ∈ N
d)
none of these
|
Geetika Nambiar answered |
I'm sorry, I cannot answer without a complete statement or context. Can you please provide more information?
If n ∈ N then n3+2n is divisible by- a)3
- b)2
- c)6
- d)4
Correct answer is option 'A'. Can you explain this answer?
If n ∈ N then n3+2n is divisible by
a)
3
b)
2
c)
6
d)
4
|
Anjali Deshpande answered |
Induction works in the following way: If you show that the result being true for any integer implies it is true for the next, then you need only show that it is true for n=1 for it to be true for n=2 and then n=3 and so on.
Step 1: Show true for n=1
For n=1, n^3+2n=(1)^3+2(1)
n^3+2n=3
3 is definitely divisible by 3 so the statement is true for n=1.
Step 2: Assume true for n=k
We assume that for any integer k, n^3+2n is divisible by 3. We can write this mathematically as:
k^3+2k=3m, where m is an integer
Step 3: Show true for k+1
For n=k+1,
n^3+2n=(k+1)^3+2(k+1)
=(k^3+3k^2+3k+1)+2k+2
=(k^3+2k)+3(k^2+k+1)
Subbing in from part 2 for (k^3+2k), we get:
n^3+2n=3m+3(k^2+k+1)
=3(m+k^2+k+1)
which is divisible by 3.
This means that the statement being true for n=k implies the statement is true for n=k+1, and as we have shown it to be true for n=1 the proof of the statement follows by induction.
The smallest +ve integer n , for whichn! 
- a)3
- b)1
- c)2
- d)4
Correct answer is option 'C'. Can you explain this answer?
The smallest +ve integer n , for whichn!
a)
3
b)
1
c)
2
d)
4
|
Amrit Raj answered |
Put the Value of n=1,2,3
for n= 1 both Side answer is 1
för n=2 Left Side answer is 2 but Right Side answer is 2.25
for n= 1 both Side answer is 1
för n=2 Left Side answer is 2 but Right Side answer is 2.25
x(xn-1 - nan-1)+ an (n-1) is divisible by (x-a)2 for- a)n > 1
- b)n > 2
- c)all n ∈ N
- d)none of these.
Correct answer is option 'C'. Can you explain this answer?
x(xn-1 - nan-1)+ an (n-1) is divisible by (x-a)2 for
a)
n > 1
b)
n > 2
c)
all n ∈ N
d)
none of these.
|
|
Maitri Das answered |
Solution:
Given Expression:
$$x(x^{n-1} - a^{n-1}) + a^n (n-1)$$
Proof:
Using Factor Theorem:
- We know that if a polynomial f(x) is divisible by (x-a), then f(a) = 0.
- Let's substitute x = a in the given expression.
Substitute x = a:
- $$a(a^{n-1} - a^{n-1}) + a^n (n-1)$$
- $$a(0) + a^n (n-1)$$
- $$a^n (n-1)$$
Conclusion:
- Since the expression becomes a multiple of (x-a) when x = a, it is divisible by (x-a)^2 for all natural numbers n.
- Therefore, the given expression is divisible by (x-a)^2 for all n ∈ N.
Correct Answer: c) all n ∈ N
Given Expression:
$$x(x^{n-1} - a^{n-1}) + a^n (n-1)$$
Proof:
Using Factor Theorem:
- We know that if a polynomial f(x) is divisible by (x-a), then f(a) = 0.
- Let's substitute x = a in the given expression.
Substitute x = a:
- $$a(a^{n-1} - a^{n-1}) + a^n (n-1)$$
- $$a(0) + a^n (n-1)$$
- $$a^n (n-1)$$
Conclusion:
- Since the expression becomes a multiple of (x-a) when x = a, it is divisible by (x-a)^2 for all natural numbers n.
- Therefore, the given expression is divisible by (x-a)^2 for all n ∈ N.
Correct Answer: c) all n ∈ N
Let P (k) = 1 + 3 + 5 + …………….+ (2k – 1) = (3+k2).Then which of the following is true ?- a)P (k) ⇒ P (k + 2)
- b)P (1) is correct
- c)Principal of mathematical induction can be used to prove the formula
- d)P (k) ⇒ P (k+1)
Correct answer is option 'D'. Can you explain this answer?
Let P (k) = 1 + 3 + 5 + …………….+ (2k – 1) = (3+k2).Then which of the following is true ?
a)
P (k) ⇒ P (k + 2)
b)
P (1) is correct
c)
Principal of mathematical induction can be used to prove the formula
d)
P (k) ⇒ P (k+1)
|
|
Saranya Chavan answered |
There seems to be some missing information in the question. The sequence P(k) is given as 1, 3, 5, but it is not clear what the pattern is or what is being asked. Please provide more context or details to the question.
A student was asked to prove a statement P (n) by method of induction. He proved P (k + 1) is true whenever P (k) Is true for all k ≥ 5 , k ∈ N and P (5) is true. On the basis of this he could conclude that P (n) is true- a)for all n > 5
- b)for all n < 5
- c)for all n ≥ 5
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
A student was asked to prove a statement P (n) by method of induction. He proved P (k + 1) is true whenever P (k) Is true for all k ≥ 5 , k ∈ N and P (5) is true. On the basis of this he could conclude that P (n) is true
a)
for all n > 5
b)
for all n < 5
c)
for all n ≥ 5
d)
none of these
|
|
Pranav Datta answered |
WKT
P(k + 1) is true whenever P(k) is true for all k ≥ 5, k ∈ N and P (5) is true.
P(n) is true for all n ≥ 5
P(k + 1) is true whenever P(k) is true for all k ≥ 5, k ∈ N and P (5) is true.
P(n) is true for all n ≥ 5
If xn−1 is divisible by x – k for all n belongs to natural numbers N, then the least positive integral value of k is :- a)2
- b)1
- c)3
- d)4
Correct answer is option 'B'. Can you explain this answer?
If xn−1 is divisible by x – k for all n belongs to natural numbers N, then the least positive integral value of k is :
a)
2
b)
1
c)
3
d)
4
|
|
Sreemoyee Gupta answered |
I'm sorry, but I'm not familiar with the term "Ifxn." Could you please provide more context or clarify your question?
A student was asked to prove a statement P (n) by method of induction. He proved that P (3) is true and that P(n) ⇒ P(n+1) for all natural numbers n. On the basis of this he could conclude that P (n) is true- a)for no n
- b)for all n ≥ 3
- c)for all n ∈ N
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A student was asked to prove a statement P (n) by method of induction. He proved that P (3) is true and that P(n) ⇒ P(n+1) for all natural numbers n. On the basis of this he could conclude that P (n) is true
a)
for no n
b)
for all n ≥ 3
c)
for all n ∈ N
d)
none of these
|
|
Naina Bansal answered |
A student was asked to prove a statement P(n) by method of induction. He proved that P(3) is true and that P(n) implies P(n+1) for all natural numbers n. On the basis of this, he could conclude that P(n) is true for all n >= 3.
The principle of mathematical induction states that if we can show that a statement P(n) is true for a particular base case (such as n = 3) and that the statement P(n) implies P(n+1) for all natural numbers n, then we can conclude that the statement P(n) is true for all natural numbers n.
Since the student was able to show that P(3) is true and that P(n) implies P(n+1) for all natural numbers n, he can conclude that P(n) is true for all n >= 3.
The correct answer is therefore (b) for all n ≥ 3.
To prove a formula by principle of mathematical induction, we verify the statement- a)For as many positive integral values of n as we like
- b)only for n = k, where k is any positive integer.
- c)Once the formula is proved for a particular positive integer the formula will automatically follow for the next positive integer and the next indefinitely.
- d)only for n = 1
Correct answer is option 'B'. Can you explain this answer?
To prove a formula by principle of mathematical induction, we verify the statement
a)
For as many positive integral values of n as we like
b)
only for n = k, where k is any positive integer.
c)
Once the formula is proved for a particular positive integer the formula will automatically follow for the next positive integer and the next indefinitely.
d)
only for n = 1
|
|
Siddharth Mehra answered |
In proof by mathematical induction, if P(k) is true then P(k+1) is also true.
32n+2−8n−9is divisible by 64 for all- a)n ∈ N
- b)n ∈ N , n > 2
- c)n ∈ N , n ≥ 2
- d)none of these.
Correct answer is option 'A'. Can you explain this answer?
32n+2−8n−9is divisible by 64 for all
a)
n ∈ N
b)
n ∈ N , n > 2
c)
n ∈ N , n ≥ 2
d)
none of these.
|
Pranavi Iyer answered |
32n+2−8n−9=9n+1−9−8n32n+2−8n−9=9n+1−9−8n
=9(9n−1)−8n=9(9n−1)−8n
=9((8+1)n−1)−8n=9((8+1)n−1)−8n
=9∑k=1n(nk)8k−8n=9∑k=1n(nk)8k−8n
==9∑k=2n(nk)8k+64n9∑k=2n(nk)8k+64n
is a multiple of 64
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