Test: Applications Of Mathematical Induction - Commerce MCQ

For n=1, we have 7² +16 -1 = 64 which is divisible by 64
 
For n=k, we assume that 7^2k + 16k-1 is divisible by 64
 
We must check for n=k+1, is the statement true
7^2(k+1) +16(k+1) -1 = 49 * 7^2k + 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*7^2k +16k +15 = 49*7^2k + 49*16k -49 -49*16k -49 +16k +15 =
49(7^2k +16k -1) - 49(16k-1) +16k +15 =
49(7^k +16k -1) - 48*16k +64
all these terms are divisible by 64

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)}

Now, for n = 1, 2*4⁽²⁺¹⁾ + 3⁽³⁺¹⁾ = 2*4³ + 3⁴ 
= 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

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.

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 = {(a^2, 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)}

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.

 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.

 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,

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): 2^3k – 1 is divisible by 7.
or  2^3k -1 = 7m, m∈ N  (i)
Now, we have to prove that P(k + 1) is true.
P(k+ 1): 2³(k+1)– 1
= 2^3k.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.

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

(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 

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

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.