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Test: Applications Of Mathematical Induction - JEE MCQ


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15 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Applications Of Mathematical Induction

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Test: Applications Of Mathematical Induction - Question 1

For each n ∈ N, 72n + 16n – 1 is divisible by :

Detailed Solution for Test: Applications Of Mathematical Induction - Question 1

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

Test: Applications Of Mathematical Induction - Question 2

 If  then A10 equals

Detailed Solution for Test: Applications Of Mathematical Induction - Question 2

A = {(1,x) (0,1)}
A2 = {(1,x) (0,1)} {(1,x) (0,1)}
= {(1+0, x+x) (0, 1)}
= {(1, 2x) (0,1)}
Similarly A10 = {(1, 10x) (0,1)}

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Test: Applications Of Mathematical Induction - Question 3

2.42n + 1 + 33n + 1 for all n is divisible by λ for all n ∈ N” is true, then the value of λ is ____

Detailed Solution for Test: Applications Of Mathematical Induction - Question 3

Now, for n = 1, 2*4(2+1) + 3(3+1) = 2*43 + 3
= 2*64 + 81 = 128 + 81 = 209, for n = 2,  2*45 + 37 
= 8*256 + 2187
= 2048 + 2187 = 4235 
Note that the H.C.F. of 209 and 4235 is 11. 
So, 2*4(2n+1) + 3(3n+1) is divisible by 11. 
Hence, λ is 11

Test: Applications Of Mathematical Induction - Question 4

The sum of cubes of three consecutive natural numbers is divisible by:

Detailed Solution for Test: Applications Of Mathematical Induction - Question 4

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.

Test: Applications Of Mathematical Induction - Question 5

If then for some n ∈ N, An is equal to :

Detailed Solution for Test: Applications Of Mathematical Induction - Question 5

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)}

Test: Applications Of Mathematical Induction - Question 6

The smallest positive integer n, for which 

Detailed Solution for Test: Applications Of Mathematical Induction - Question 6

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.

Test: Applications Of Mathematical Induction - Question 7

To prove a formula by principle of mathematical induction, we verify the statement

Detailed Solution for Test: Applications Of Mathematical Induction - Question 7

 Let P(n) be a given statement involving the natural number n such that
(i) The statement is true for n = 1, i.e., P(1) is true (or true for any fixed natural
number) and
(ii) If the statement is true for n = k (where k is a particular but arbitrary natural
number), then the statement is also true for n = k + 1, i.e, truth of P(k) implies
the truth of P(k + 1). Then P(n) is true for all natural numbers n.

Test: Applications Of Mathematical Induction - Question 8

2n > n2 when n ∈ N such that

Detailed Solution for Test: Applications Of Mathematical Induction - Question 8

 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,

Test: Applications Of Mathematical Induction - Question 9

The greatest natural number, which divides (n + 1) (n +2) (n + 3)(n + 4) is

Test: Applications Of Mathematical Induction - Question 10

If P(k) is the statement 23k – 1 is divisible by 7, then P(k + 1) is

Detailed Solution for Test: Applications Of Mathematical Induction - Question 10

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.

Test: Applications Of Mathematical Induction - Question 11

For each n ∈ N,  n(n + 1) (2n + 1) is divisible by:

Test: Applications Of Mathematical Induction - Question 12

Let P(n) is the statement the product of two consecutive natural numbers is even. Then which of the following is true?

Detailed Solution for Test: Applications Of Mathematical Induction - Question 12

The product of two consecutive natural number is natural.
And all even number be natural number.
So, P(n), n ∈ N

Test: Applications Of Mathematical Induction - Question 13

For positive integers n, (1 + x)n > 1 + nx ; (x > – 1) when

Detailed Solution for Test: Applications Of Mathematical Induction - Question 13

(1 + x)n > 1 + nx
For n = 1, (1 + x)n = (1 + x)1 = 1 + x
1 + nx = 1 + x
(1 + x)n = 1 + nx        
For n = 2, (1 + x)n = (1 + x)2 = 1 + x2 + 2x
1 + nx = 1 + 2x
(1 + x)n ≥ 1 + nx (Equality when x = 0)
So, (1 + x) n > 1 + nx for n ≥ 2 

Test: Applications Of Mathematical Induction - Question 14

Let P(n) b e a statement 2n<n! where n is a natural number, then P(n) is true for

Detailed Solution for Test: Applications Of Mathematical Induction - Question 14

P(n)  b e a statement 2n < n!
for(n=1) 21 < 1!
=> 2 < 1(false)
for(n=2) 22 < 2!
=> 4 < 2 (false)
for(n=3) 23 < 3!
=> 8 < 6(false)
for(n=4) 24 < 4!
=> 16 < 24 (true)
for(n=5) 25 < 5!
=> 32 < 120 (true)
Therefore 2n < n! is greater for all n>3

Test: Applications Of Mathematical Induction - Question 15

The principle of mathematical induction is for the set of:

Detailed Solution for Test: Applications Of Mathematical Induction - Question 15

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.

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