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The first principle of mathematical induction states that if the basis step and the inductive step are proven, then P(n) is true for all natural numbers.

Let P(n):c(a₁+a₂+…+an) = ca₁+ca₂+…+ca_n ,for all natural 

numbers, n ≥ 2. 

Step1: For n=2, 

P(2) 

LHS= c(a₁ + a₂) 

RHS= c a₁ + ca₂ 

As, it is given that c(a₁ + a₂) = c a₁ + ca₂ 

Thus, P(2) is true. 

Step2: For n=k, 

Let P(k) be true 

So, c(a₁+a₂+…+a_k ) = ca₁+ca₂+…+ca_k 

Now, we need to show P(k+1) is true whenever P(k) is true.

P(k+1): 

LHS= c(a₁+a₂+…+a_K+a_k+1) 

=c[(a₁+a₂+…+a_K)+a_k+1] 

=c(a₁+a₂+…+a_K)+ca_k+1 

=ca₁+ca₂+…+ca_K+ca_k+1 

= RHS 

Thus, P(k+1) is true, so by mathematical induction P(n) is true.

Let P(n): n(n + 1) (n + 5) is multiple by 3 for all n∈N 

Let P(n) is true for n=1 

P(1): 1(1 + 1) (1 + 5) 

= 2 × 6 

= 12 

Since, it is multiple of 3 

So, P(n) is true for n = 1 

Now, Let P(n) is true for n = k 

P(k): k(k + 1) (k + 5) 

= k(k + 1) (k + 5) is a multiple of 3 

Then, k(k + 1) (k + 5) = 3λ - - - - - (1)

We have to show, 

= (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3 

= (k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ 

Now, 

= (k + 1)[(k + 1) + 1][(k + 1) + 5] 

= (k + 1)(k + 2)[(k + 1) + 5] 

= [k(k + 1) + 2(k + 1)][(k + 5) + 1] 

= k(k + 1)(k + 5) + k(k + 1) + 2(k + 1)(k + 5) + 2(k + 1) 

= 3λ + k² + k + 2(k² + 6k + 5) + 2k + 2 

= 3λ + k² + k + 2k² + 12k + 10 + 2k + 2 

= 3λ + 3k 2 + 15k + 12 

= 3(λ + k 2 + 5k + 4) 

= 3μ 

Therefore, P(n) is true for n = k + 1 

Hence, P(n) is true for all n∈N

For n=1, 

49 16 + λ 

⇒ 65 + λ 

Now we can see that if λ = -1, then it is divisible by 64 

λ = - 1

Let us considering P (n) = 1² + 2² + 3² + … + n² = [n(n + 1) (2n + 1)]/6

For, n = 1

P (1) = [1(1 + 1) (2 + 1)]/6

1 = 1

P (n) is true for n = 1

Suppose P (n) is true for n = k, therefore

P (k): 1² + 2² + 3² + … + k² = [k(k + 1) (2k + 1)]/6

Let’s us check for the P (n) = k + 1, therefore

P(k) = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = [k + 1 (k + 2) (2k + 3)]/6  

= 1² + 2² + 3² + – – – – – + k² + (k + 1)²

= [k + 1 (k + 2) (2k + 3)]/6 + (k + 1)²

= (k +1) [(2k² + k)/6 + (k + 1)/1]

= (k +1) [2k² + k + 6k + 6]/6

= (k +1) [2k² + 7k + 6]/6

= (k +1) [2k² + 4k + 3k + 6]/6

= (k +1) [2k(k + 2) + 3(k + 2)]/6

= [(k +1) (2k + 3) (k + 2)] / 6

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Suppose P (n): (ab)ⁿ = aⁿ bⁿ 

Now let us check for n = 1,

P (1): (ab)¹ = a¹ b¹

: ab = ab

P (n) is true for n = 1.

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): (ab)^k = a^k b^k … (i)

Now we have to prove,

(ab)^{k + 1} = a^{k + 1}.b^{k + 1}

Therefore,

= (ab)^{k + 1}

= (ab)^k (ab)

= (a^k b^k) (ab) using equation (1)

= (a^{k + 1}) (b^{k + 1})

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Let us considering

P (n) = 1 + 2 + 3 + – – – – – + n = n(n+1)/2 

Therefore,

P (n) is true for all natural numbers.

Thus, P (n) is true for all n ∈ N.

Let us considering P(n) = 1 + 2 + 3 + ….. + n = n(n +1)/2

For the n = 1

LHS of P (n) = 1

RHS of P (n) = 1 (1+1)/2 = 1

Therefore, LHS = RHS

Here, P (n) is true for n = 1

Let us consider P (n) be the true for n = k, therefore

1 + 2 + 3 + …. + k = k (k+1)/2 … (i)

Then,

(1 + 2 + 3 + … + k) + (k + 1) = k(k + 1)/2 + (k + 1)

= (k + 1) (k/2 + 1)

= [(k + 1) (k + 2)]/2

= [(k + 1) [(k + 1) + 1]]/2

P (n) is true for n = k + 1

P (n) is true for all n ∈ N

Therefore, by the principle of Mathematical Induction

Thus, P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2 is true for all n ∈ N.

Suppose P (n) = 1 + 3 + 3² + – – – – + 3^{n – 1} = (3ⁿ – 1)/2 

Then, For n = 1

P (1) = 1 = (3¹ – 1)/2 = 2/2 =1

P (n) is true for n = 1

Then, let’s us check for the P (n) is true for n = k

P (k) = 1 + 3 + 3² + – – – – + 3^{k – 1} = (3^k – 1)/2 … (i) 

Then, we have to show P (n) is true for n = k + 1

P (k + 1) = 1 + 3 + 3² + – – – – + 3^k = (3^{k + 1} – 1)/2 

Now, {1 + 3 + 3² + – – – – + 3^{k – 1}} + 3^{k + 1 – 1}

= (3k – 1)/2 + 3^k using equation (i)

= (3k – 1 + 2 × 3^k)/2

= (3 × 3 k – 1)/2

= (3^{k + 1} – 1)/2

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Given as

P(n) = n² – n + 41 is prime.

P(n) = n² – n + 41

P(1) = 1 – 1 + 41

= 41

P (1) is Prime.

Now similarly,

P(2) = 2² – 2 + 41

= 4 – 2 + 41

= 43

P (2) is prime.

Then similarly,
P (3) = 3² – 3 + 41

= 9 – 3 + 41

= 47

P (3) is prime

Then,

P (41) = (41)² – 41 + 41

= 1681

P (41) is not prime

Thus P (1), P(2), P (3) are true but P (41) is not true.

Let M be an integer. Suppose we want to prove that P(n) is true for all positive integers ≥M. Then if we show that: 

Step 1: P(M) is true, and 

Step 2: for an arbitrary positive integer k≥M, if P(M).P(M+1).P(M+2)……P(k) are true then P(k+1) is true, 

Then P(n) is true for all positive integers greater than or equal to M.

Suppose P (n): n(n + 1) (n + 5) is a multiple of 3

Now let us check for n = 1,

P(1): 1(1 + 1) (1 + 5)

: 2 × 6

: 12

P (n) is true for n = 1. Where, P (n) is a multiple of 3

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): k(k + 1) (k + 5) is a multiple of 3

: k(k + 1) (k + 5) = 3λ … (i)

Now we have to prove,

(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

Therefore,

= (k + 1) [(k + 1) + 1] [(k + 1) + 5]

= (k + 1) (k + 2) [(k + 1) + 5]

= [k(k + 1) + 2(k + 1)] [(k + 5) + 1]

= k(k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)

= 3λ + k² + k + 2(k² + 6k + 5) + 2k + 2

= 3λ + k² + k + 2k² + 12k + 10 + 2k + 2

= 3λ + 3k² + 15k + 12

= 3(λ + k² + 5k + 4)

= 3μ

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n = \(\frac{n(n+1)}2\)

For n = 1 

L.H.S of P(n) = 1 

R.H.S of P(n) = \(\frac{1(1+1)}2\) = \(\frac{2}2\) = 1 

Therefore, L.H.S = R.H.S 

Since, P(n) is true for n = 1 

Let assume P(n) be the true for n = k, 

so 

1 + 2 + 3 + - - - - - + k = \(\frac{k(k+1)}2\) - - - (1) 

Now 

(1 + 2 + 3 + - - + k) + (k + 1) 

= \(\frac{k(k+1)}2\) + (k + 1)

=  (k + 1)\((\frac{k}2+1)\)

= \(\frac{(k+1)(k+2)}2\)

=  \(\frac{(k+1)[(k+1)+1]}2\)

P(n) is true for n = k + 1 

P(n) is true for all n∈N 

So , by the principle of Mathematical Induction

Hence, P(n) = 1 + 2 + 3 + - - - + n = \(\frac{n(n+1)}2\) is true for all n∈N

Suppose P (n) = 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)

For, n = 1

P (n) = 1/1.2 = 1/1+1

1/2 = 1/2

P (n) is true for n = 1

Now, let’s check for P (n) is true for n = k,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k+1) + k/(k+1) (k+2) = (k+1)/(k+2)

Now,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k + 1) + k/(k + 1) (k + 2)

= 1/(k + 1)/(k + 2) + k/(k + 1)

= 1/(k + 1) [k(k + 2) +1]/(k + 2)

= 1/(k + 1) [k² + 2k + 1]/(k + 2)

=1/(k + 1) [(k + 1) (k + 1)]/(k + 2)

= (k + 1)/(k + 2)

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Suppose P (n): 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25

Now let us check for n = 1,

P (1): 7² + 2⁰.3⁰

: 49 + 1

: 50

P (n) is true for n = 1. Where, P (n) is divisible by 25

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 7^2k + 2^{3k – 3}. 3k – 1 is divisible by 25

: 7^2k + 2^{3k – 3}. 3^{k – 1} = 25λ … (i)

Now we have to prove that:

7^{2k + 1} + 2^3k. 3^k is divisible by 25

7^{2k + 2} + 2^3k. 3^k = 25μ

Therefore,

= 7^{2(k + 1)} + 2^3k. 3^k

= 7^2k.7¹ + 2^3k. 3^k

= (25λ – 2^{3k – 3}. 3^{k – 1}) 49 + 2^3k. 3k by using equation (i)

= 25λ. 49 – 2^3k/8. 3^k/3. 49 + 2^3k. 3^k

= 24 × 25 × 49λ – 2^3k . 3^k . 49 + 24 . 2^3k.3^k

= 24 × 25 × 49λ – 25 . 2^3k. 3^k

= 25(24 . 49λ – 2^3k. 3^k)

= 25μ

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

for n=1, 

2×4^2×1+1 + 3^3×1+1=2×4³+3⁴ 

= 2×64+81 

= 128+81 

= 209 

For n=2, 

2×4^2×2+1 + 3^3×2+1 = 2×4⁵+3⁷ 

= 2×1024+2187 

= 2048+2187 

= 4235 

Now, the H.C.F of 209 and 4235 is 11. 

Hence, λ=11.

Let the given statement be defined as 

P(n): The number of subsets of a set containing n distinct 

elements = 2ⁿ, for all n ϵ N. 

Step1: For n = 1, 

L.H.S=As, the subsets of the set containing only 1 element are: 

Φ and the set itself 

i.e. the number of subsets of a set containing only element = 2 

R.H.S = 2¹ = 2 

As, LHS=RHS, so, it is true for n=1. 

Step2: Let the given statement be true for n = k. 

P(k): The number of subsets of a set containing k distinct 

elements = 2^k 

Now, we need to show P(k+1) is true whenever P(k) is true. 

P(k+1): 

Let A={a₁, a₂, a₃, a₄,…, a_k , b} so that A has (k+1) elements. 

So the subset t of A can be divided into two collections: 

first contains subsets of A which don t have b in them and the second contains subsets of A which do have b in them. 

First collection: { }, {a₁},{a₁, a₂},{a₁, a₂, a₃},…,{a₁, a₂, a₃, a₄,…, a_k} and 

Second collection: {b}, {a₁,b},{a₁,a₂,b },{a₁,a₂,a₃,b},…,{a₁,a₂,a₃,a₄,…,a_k , b} 

It can be clearly seen that: 

The number of subsets of A in first collection 

= The number of subsets of set with k elements i.e. {a₁, a₂, a₃, a₄,…, ak} = 2^k 

Also it follows that the second collection must have 

the same number of the subsets as that of the first = 2^k 

So the total number of subsets of A = 2^k+2^k = 2^k+1 

Thus, by the principle of mathematical induction P(n) is true.

P(n) = n² - n + 41 

= P(1) = 1 - 1 + 41 

= P(1) = 41 

Therefore, P(1) is Prime 

= P(2) = 2² – 2 + 41 

= P(2) = 4 - 2 + 41 

= P(2) = 43 

Therefore, P(2) is prime 

= P(3) = 3² – 3 + 41 

= P(3) = 9 – 3 + 41 

= P(3) = 47 

Therefore P(3) is prime 

Now, P(41) = (41)² - 41 + 41 

= P(41) = 1681 

Therefore, P(41) is not prime 

Hence, P(1),P(2),P(3) are true but P(41) is not true.

Suppose P (n): 1 + 3 + 5 + … + (2n – 1) = n²

Now, let us check for the P (n) is true for n = 1

P (1) = 1 =1²

1 = 1

P (n) is true for n = 1

Then, let’s us check for the P (n) is true for n = k

P (k) = 1 + 3 + 5 + … + (2k – 1) = k² … (i)

Now, we have to show that

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)²

Then,

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

= k² + (2k + 1)

= k² + 2k + 1

= (k + 1)²

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

P(n) = 1 + 2 + 3 + - - - - - + n = \(\frac{n(n+1)}2\)

P(n) is true for all natural numbers. 

Hence, P(n) is true for all n∈N

If P(r) is true then 2^r ≥ 3r 

For, P(r+1) 

2^r+1 = 2.2^r 

For, x>3, 2x>x+3 

So, 2.2^r > 2^r + 3 for r >1 

⇒ 2^r+1>2^r+3 for r>1 

⇒ 2^r+1 > 3r +3 for r>1 

⇒ 2^r+1 > 3(r+1) for r>1 

So, if P(r) is true, then P(r+1) is also true. 

For, n =1, P(1): 

L.H.S = 2 

R.H.S = 3 

As L.H.S < R.H.S

So, it is not true for n = 1 

Hence, P(n) is not true for all natural numbers.

Given as 

P (n) = “2ⁿ ≥ 3n” and p(r) is true.

We have, P (n) = 2ⁿ ≥ 3n

Here, P (r) is true

Therefore,

2^r ≥ 3r

Then, let’s multiply both sides by 2

2 × 2^r ≥ 3r × 2

2^{r + 1} ≥ 6r

2^{r + 1} ≥ 3r + 3r [since 3r >3 = 3r + 3r ≥ 3 + 3r]

∴ 2^{r + 1} ≥ 3(r + 1)

Thus, P (r + 1) is true.

Given for all n€ N 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ 

For n=1, 

3 × 5³ + 2⁴ 

3 × 125 + 16 

375 + 16 = 391 

For n=2, 

3 × 5⁵ + 2⁷ 

3 × 3125 +128 

9375 + 128 = 9503 

H.C.F of 391, 9503 = 17

Answer: (B) S(k) ⇒ S(k+1)

S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k² 

Putting k = 1 on both the sides, we get 

LHS = 1, RHS = 3 + 1 = 4 

⇒ LHS ≠ RHS 

⇒ S(1) is not true. 

Assume S(k) = 1 + 3 + 5 + .... + (2k – 1) = 3 + k² is true. 

Then, 

To find S(k + 1), add the (k + 1)^th term = (2 (k + 1) – 1) = 2k +1 on both the sides of S(k). 

∴ S(k + 1) = 1 + 3 + 5 + .... + (2k – 1)+(2k + 1) = 3 + k² + 2k + 1 

⇒ 1 + 3 + 5 + .... + (2k – 1) + (2k + 1) = 3 + (k + 1)² 

⇒ S(k + 1) is also true. 

∴ S(k) ⇒ S(k + 1) is true

Answer: (C) n(n+1)

Let S_n = 2 + 4 + 6 + .... + 2n 

When n = 1, S_n = 2 

Now from the options given, when n = 1, 

2(n + 1) = 4, \(\frac{1}{2}n(n+2) =\frac{3}{2}\) ,n(n + 1) = 2, (n + 2) (n + 4) = 15

∴ Sn ≠ 2(n + 1), Sn ≠ \(\frac{1}{2}\) n (n + 2), Sn ≠ (n + 2) (n + 4) for n = 1 

Sn = n (n + 1) for n = 1 

∴  We need to prove 2 + 4 + 6 + .... + 2n = n (n + 1) ∀ n∈N. 

Let T(n) = 2 + 4 + 6 + .... + 2n = n (n + 1) 

Basic Step: 

For n =1, LHS = 2 × 1 = 2, RHS = 1 × (1 + 1) = 2 

⇒ LHS = RHS 

⇒ T(1) is true.

Induction Step: 

Assume T(k) is true, i.e., 

2 + 4 + 6 + .... + 2k = k (k + 1) 

To obtain T(k + 1), we add the (k + 1)^th term, i.e, 2 (k + 1) to both the sides of T(k), 

i.e., 2 + 4 + 6 + ... + 2k + 2(k + 1) = k(k + 1) + 2(k + 1) 

= (k + 1) (k + 2) = (k + 1) ((k + 1) + 1) 

Thus the statement T(n) is true for n = k + 1, whenever it is true for n = k. 

Therefore by the principle of mathematical induction it is true for all n∈N

Given 

P(r) is true that means, 

= r² + r is even 

Let Assume r² + r = 2k - - - - - - (i) 

Now, (r + 1)² + (r + 1) 

r² + 1 + 2r + r + 1 

= (r² + r) + 2r + 2 

= 2k + 2r + 2 

= 2(k + r + 1) 

= 2μ 

Therefore, (r + 1)² + (r + 1) is Even. 

Hence, P(r + 1) is true

Suppose P (n): 3²ⁿ + 7 is divisible by 8 

Now let us check for n = 1,

P (1): 3² + 7 = 9 + 7 = 16

P (n) is true for n = 1. Where, P (n) is divisible by 8

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 3^2k + 7 is divisible by 8

: 3^2k + 7 = 8λ

: 3^2k = 8λ – 7 … (i)

Now we have to prove,

3^{2(k + 1)} + 7 is divisible by 8

3^{2k + 2} + 7 = 8μ

Therefore,

= 3^{2(k + 1)} + 7

= 3^2k.3² + 7

= 9.3^2k + 7

= 9.(8λ – 7) + 7 by using equation (i)

= 72λ – 63 + 7

= 72λ – 56

= 8(9λ – 7)

= 8μ

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Given. 

P(n) = “2ⁿ ≥ 3n” and p(r) is true. 

Prove. P(r + 1) is true 

we have P(n) = 2ⁿ ≥ 3n 

Since, P(r) is true So, 

= 2^r ≥ 3r 

Now, Multiply both side by 2 

= 2.2^r ≥ 3r.2 

= 2^r+1 ≥ 6r 

= 2^r+1 ≥ 3r + 3r [since 3r >3 = 3r + 3r ≥3 + 3r] 

Therefore 2^r+1 ≥ 3(r + 1) 

Hence, P(r + 1) is true

Given P(n):2n< (1×2×…. ×n)

For n=1, 2<2 

For n=2, 4<4 

For n=3, 6<6 

For n=4, 8<24 

∴ the smallest positive integer for which P(n) is true is 4.

Answer: (B) an integral multiple of 6 

When n = 1, n(n+1)(2n+1) = (1) (2) (3) = 6, which is an integral multiple of 6. 

It is neither an odd integer nor a perfect square. 

Using the principle of mathematical induction, we shall now show that the expression n(n + 1)(2n + 1) is an integral multiple of 6 ∀ n ∈ N. 

Assume T(n) = n(n+1)(2n+1) = 6x where x∈N. 

Basic Step: 

T(1) is true as shown above. 

Induction Step: 

Let T(k) be true for all k∈N. 

⇒ k (k +1) (2k +1) = 6x, where x∈N. .....(i) 

For T(k + 1), we replace k by (k + 1) in the given expression, i.e., 

T(k + 1) = (k + 1) (k + 2) (2 (k + 1) + 1) 

= (k + 1) (k + 2) ((2k + 1) + 2) 

= (k + 1) (k + 2) (2k + 1) + 2 (k + 1) (k + 2) 

= k (k + 1) (2k + 1) + 2 (k + 1) (2k + 1) + 2 (k + 1) (k + 2) 

= k (k + 1) (2k + 1) + 2 (k + 1) [(2k + 1) + (k + 2)] 

= 6x + 2(k + 1)(3k + 3) 

= 6x + 6(k + 1)² 

= 6 (x + (k + 1)²) = 6 × a positive integer 

∴ T(k) is true ⇒ T(k + 1) is true. 

∴ T(n) is true for all n∈N

Suppose P (n): 5^{2n + 2} – 24n – 25 is divisible by 576

Now let us check for n = 1,

P (1): 5^{2.1 + 2} – 24.1 – 25

: 625 – 49

: 576

P (n) is true for n = 1. Where, P (n) is divisible by 576

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): 5^{2k + 2} – 24k – 25 is divisible by 576

: 5^{2k + 2} – 24k – 25 = 576λ …. (i)

Now we have to prove,

5^{2k + 4} – 24(k + 1) – 25 is divisible by 576

5^{(2k + 2) + 2} – 24(k + 1) – 25 = 576μ

Therefore,

= 5^{(2k + 2) + 2} – 24(k + 1) – 25

= 5^{(2k + 2)}.5² – 24k – 24 – 25

= (576λ + 24k + 25)25 – 24k– 49 by using equation (i)

= 25. 576λ + 576k + 576

= 576(25λ + k + 1)

= 576μ

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Let P(n) = 1×1! + 2×2! + 3×3! +…+ n×n 

P(n): 1×1! + 2×2! + 3×3! +…+ n×n! = (n + 1)! – 1 for all nϵ N 

Step1: 

P(1) = 1×1! = (2)! – 1 = 1 

Thus, P(n) is equal to (n + 1)! – 1 for n = 1 

Step2: 

Let, P(m) be equal to (m + 1)! – 1 

Then, 1×1! + 2×2! + 3×3! +…+ m×m! = (m + 1)! – 1 

Now, we need to show that P(m+1) is true whenever P(m) is true. 

P(m+1) = 1×1! + 2×2! + 3×3! +…+ m×m! + (m+1)×(m+1)!

= (m+1)! – 1 + (m+1)×(m+1)! 

= (m+1)!(m+1+1) – 1 

= (m+1)!(m+2) – 1 

= (m+2)! – 1 Thus, P(m+1) is true. 

So, by the principle of mathematical induction, P(n) is true for all nϵN.

Let P(n) = 2.7ⁿ + 3.5ⁿ – 5 

Now, P(n): 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n ϵ N 

Step1: 

P(1) = 2.7 + 3.5 – 5 = 1.2 

Thus, P(1) is divisible by 24 

Step2: 

Let, P(m) be divisible by 24 

Then, 2.7^m + 3.5^m – 5 = 24λ, where λ ϵ N. 

Now, we need to show that P(m+1) is true whenever P(m) is true. 

So, P(m+1) = 2.7^m+1 + 3.5^m+1 – 5 

= 2.7^m+1 + 5.( 2.7^m + 3.5^m – 5 ) – 5 

= 2.7^m+1 + 5.( 24λ + 5 - 2.7^m ) – 5 

= 2.7^m+1 + 120λ + 25 - 10.7^m – 5 

= 2.7^m.7 - 10.7^m+ 120 λ +24 – 4 

= 7^m(14 – 10) + 120 λ +24 – 4 

= 7^m(4) + 120 λ +24 – 4 

= 7^m(4) + 120 λ +24 – 4 

= 4(7^m - 1) + 24(5λ +1) 

As, 7^m – 1 is a multiple of 6 for all m ϵ N. 

So, P(m+1) = 4.6μ +24(5λ +1) 

= 24(μ +5λ +1) 

Thus, P(m+1) is true. 

So, by the principle of mathematical induction, P(n) is true for all n ϵN.

Let P(n) = n³ – 7n + 3 

Now, P(n): n³ – 7n + 3 is divisible by 3 for all n ϵ N 

Step1: 

P(1) = 1 – 7 + 3 = -3 

Thus, P(1) is divisible by 3 

Step2: 

Let, P(m) be divisible by 24 

Then, n³ – 7n + 3 = 3λ, where λ ϵ N. 

Now, we need to show that P(m+1) is true whenever P(m) is true.

So, P(m+1) = (n+1)³ – 7(n+1) + 3 

= n³+3n²+3n+1-7n-7+3 

= n³– 7n + 3 +3n²+3n+1-7 

= 3λ+3(n²+n-2) 

= 3(λ+n²+n-2) 

Thus, P(m+1) is true. 

So, by the principle of mathematical induction, P(n) is true for all n ϵN.

Given. 

P(n) = n³ + n is divisible by 3 

Find P(3) is true but P(4) is not true 

We have P(n) = n³ + n is divisible by 3 

Let’s check with P(3) 

= P(3) = 3³ + 3 

= P(3) = 27 + 3 

Therefore 

P(3) = 30, So it is divisible by 3 

Now check with P(4) 

= P(4) = 4³ + 4 

= P(4) = 64 + 4 

Therefore P(4) = 68, So it is not divisible by 3 

Hence, 

P(3) is true and P(4) is not true.

Let P(n) = 1 + 2 + 2² + … + 

P(n): 1 + 2 + 2² + … + 2ⁿ = 2^{n + 1} – 1 for all n ϵ N 

Step1: 

P(1) = 1 = (2) – 1 = 1 

Thus, P(n) is equal to 2 n + 1 – 1 for n = 1 

Step2: 

Let, P(m) be equal to 2^{m + 1} – 1 

Then, 1 + 2 + 2² + … + 2^m = 2^{m + 1} – 1 

Now, we need to show that P(m+1) is true whenever P(m) is true. 

P(m+1) = 1 + 2 + 2² + … + 2^m + 2^{m + 1} 

= 2^{m + 1} – 1 + 2^{m + 1} 

= 2.2^{m + 1} –1 

= 2^{m + 2} – 1 

Thus, P(m+1) is true. 

So, by the principle of mathematical induction, P(n) is true for all nϵN.

Answer: (B) 576

For n = 1, 

5^{2n + 2} – 24n – 25 = 5⁴ – 24 – 25 = 625 – 49 = 576 which is divisible by 576 and none of the other given alternative.

∴ To prove: 5²ⁿ⁺² – 24n – 25 is divisible by 576 using mathematical induction. 

Let T(n) be the statement: 5^{2n + 2} – 24n – 25 is divisible by 576 ∀ n∈N. 

Basic Step: 

For n = 1, T(1) = 5⁴ – 24 – 25 = 576 which is divisible by 576. 

⇒ T(1) is true. 

Induction Step: 

Assume T(k) where n = k, k∈N to be true i.e., 

T(k) = 5^{2k + 2} – 24k – 25 is divisible by 576 is true, 

i.e., 5^2k+2 – 24k – 25 = 576m, m∈N ....(i) 

∴ T(k + 1) = 5^{2(k + 1)+2} – 24 (k + 1) – 25 

= 5^{2k + 2} . 25 – 24k – 24 – 25 

= 5^{2k + 2} . 25 – 24k – 49

= 25 (5^{2k + 2} – 24k – 25) + 24. (24k) + 576 

= 25. (576m) + 576k + 576 (From (i)) 

= 576 (25m + k + 1) 

⇒ 2^{2(k + 1) + 2} – 24 (k + 1) – 25 is divisible by 576 

⇒ T(k + 1) is true, whenever T(k) is true. 

⇒ 5^{2n + 2} – 24k – 25 is divisible by 576 ∀  n∈N

Answer: (D) = 3

For n = 1, n³ + 2n = 1 + 2 = 3 which is divisible by 3 and none of the other given alternatives. 

∴ We shall prove n³ + 2n divisible by 3 for all n∈N. 

Let T(n) = n³ + 2n is divisible by 3. 

Basic Step: 

For n = 1, T(1) = n³ + 2n = 1 + 2 = 3 is divisible by 3 is true. 

Induction Step: 

Assume T(k) to be true, i.e., T(k) = k³ + 2k is divisible by 3 

= k³ + 2k = 3m, where m∈N. ...(i) 

Now we need to prove that T(k + 1) holds true, i.e., 

(k + 1)³ + 2(k + 1) is divisible by 3. 

(k + 1)³ + 2(k + 1) = k³ + 3k² + 3k +1 + 2k + 2

= (k³ + 2k) + (3k² + 3k + 3) 

= 3m + 3 (k² + k + 1) (From (i)) 

⇒ T(k + 1) = (k + 1)³ + 2 (k + 1) is divisible by 3, whenever T(k) = k³ + 2k is divisible by 3. 

⇒ n³ + 2n is divisible by 3  ∀ n∈N.

Let T(n) be the statement:

\(3^{2^n}\) – 1 is divisible by 2^{n + 2} 

Basic Step: 

For n = 1, 

\(3^{2^1}\) – 1 = 8 and 2^{n + 2} = 8 

⇒ T (1) is true 

Induction Step: 

Assume T(k) to be true, i.e., 

T(k) = \(3^{2^k}\) – 1 is divisible by 2^{k + 2} 

= \(3^{2^k}\) – 1 = m. 2^{k + 2} when m∈N ...(i) 

= \(3^{2^k}\)= m. 2^{k + 2} + 1 

Now we need to prove that T(k + 1) holds true. 

∴ \(3^{2^{k+1}}\) –1 = \(3^{2^k.2}\) – 1 

= (m . 2^{k + 2} + 1)² – 1 (using (i)) 

= m² (2^k+2)² + 2m . 2^k+2 + 1 – 1 = 2^k+2 (m² . 2^k+2 + 2m) 

⇒ T(k + 1) = \(3^{2^{k+1}}\) – 1 is divisible by 2^k+2, whenever T(k) holds. 

Thus \(3^{2^n}\) – 1 is divisible by 2^{n + 2} for all integers n \(\geq\) 1.

Let T(n) be the statement: 3²ⁿ⁺² – 8n – 9 is divisible by 64. 

Basic Step: 

For n =1, 3^2×1+2 – 8 × 1 – 9 = 81 – 17 = 64 which is divisible by 64. 

⇒ T(1) holds. 

Induction Step: 

Let T(k), k∈N hold, i.e., 

3^2k+2 – 8k – 9 is divisible by 64. 

Then, T(k + 1) = 3^{2(k+1) + 2} – 8(k + 1) – 9 = 3². 3^2k+2 – 8k – 17 

= 9 (3^2k+2 – 8k – 9) + 64k + 64 = 9. 

T (k) + 64 (k + 1) 

⇒ T(k + 1) is divisible by 64, whenever T(k) is divisible by 64. 

⇒ T(n) is true for every natural number n.

We have P(n) = n(n + 1). 

= P(3) = 3(3 + 1) 

= P(3) = 3(4) 

Hence, 

P(3) = 12, So P(3) is also Even.

Let T(n) be the statement: 

xⁿ – yⁿ is divisible by x – y. 

Basic Step:

For n = 1, x¹ – y¹ = x – y is divisible by (x – y) 

⇒ T(1) is true 

Induction Step: 

Assume that T(k) is true, i.e., for k∈N x^k – y^k is divisible by (x – y) 

Now, we prove T(k + 1) is true. 

x^k+1 – y^k+1 = x^k . x – y^k . y = x^k.x – x^k . y + x^k .y – y^k . y 

(Adding and subtracting x^k.y) 

= x^k (x – y) + y (x^k – y^k ) Since x^k(x – y) is divisible by (x – y) and (x^k – y^k ) is divisible by (x – y) (By induction step, i.e., assuming T(k) is true), therefore,

x^k+1 – y^k+1 = x^k(x – y) + y(x^k – y^k) is divisible by (x – y) ⇒ T(k + 1) is true, whenever T(k) is true. 

⇒ T(n) holds for all positive integral values of n.

Answer : (b) = 24

When n = 1, 

2.7ⁿ + 3.5ⁿ – 5 = 2.7 + 3.5 – 5 = 24 which is divisible by 24 and none of the other given alternatives. 

∴ We need to prove 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 ∀  n∈N. 

Let T(n) be the statement 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24. 

T(1) holds true as shown above. 

Assume T(k) to be true, i.e., 2.7^k + 3.5^k – 5 is divisible by 24, i.e., 

2.7^k + 3.5^k – 5 = 24m, m∈N ....(i) 

Now 2.7^{k + 1} + 3.5^{k + 1} – 5 = 2.7.7^k + 3.5.5^k – 5 

= (2.7^k + 3.5^k – 5) + 12(7^k ) + 12 (5^k ) 

= 24m + 12 (7^k + 5^k ) 

\(\big[Now\,7^k\,and\,5^k\,,k\,\in\,N\,being \,both\,odd\,,their\,sum\,is\,even. Let\,7^k\,+5^k\,=2x\,,\,x\,\in\,N\big]\)

 = 24m + 12 (2x) ; m, x∈N = 24 (m + x) 

⇒ 2.7^{k + 1} + 3.5^{k + 1} – 5 is divisible by 24 

⇒ T (k + 1) is true whenever T(k) is true, k∈N. 

⇒ 2.7ⁿ + 3.5ⁿ – 5 is divisible by 24 for all n∈N.

Let T(n) be the statement: 1 + 3 + 5 + ... + (2n – 1) = n² 

Basic Step: For n = 1, LHS = 1, RHS = 1² 

⇒ LHS = RHS ⇒ T(1) is true 

Induction Step: Assume that T(k) is true, i.e., 1 + 3 + 5 + ... + (2k – 1) = k² 

To obtain T (k + 1), add the (k + 1)^th term 

= 2 (k + 1) – 1 = 2k + 2 – 1 = 2k + 1 to both the sides. 

Then, 1 + 3 + 5 + ... + (2k – 1) + (2k + 1) = k² + 2k + 1 

⇒ 1 + 3 + 5 + ... to (k + 1) terms = (k + 1)² 

Thus the statement is true for n = k + 1 under the assumption that statement is true for n = k 

Therefore, the statement 1 + 3 + 5 + ... to n terms = n² for every positive integer n.

Answer: (C) = 9

For n = 1, 

10ⁿ + 3 (4^{n + 2}) + 5 = 10 + 3 × 4³ + 5 = 10 + 192 + 5 = 207 which is divisible by only 9 and none of the other given alternatives. 

\(\therefore\) We need to prove 10ⁿ + 3 (4^{n + 2}) + 5 is divisible by 9 ∀  n∈N. 

Let T(n) be the statement 

10ⁿ + 3(4ⁿ⁺²) + 5 is divisible by 9. 

Basic Step: 

For n = 1, T(1) holds true as proved above. 

Induction Step: 

Assume T(k) to be true, k∈N i.e., 

10^k + 3 (4^{k + 2}) + 5 is divisible by 9, i.e., 

10^k + 3 (4^{k + 2}) + 5 = 9m, m∈N ....(i) 

Now, 10^{k + 1} + 3 (4^k+1+2) + 5 

= 10^k+1 + 3(4^{k + 3}) + 5 

= 10. 10^k + 12. 4^k+2 + 5 

= 4(10^k + 3(4^{k + 2})+5) + 6 .10^k – 15 

= 4. (9m) + 6 (10^k – 1) – 9 

= 4. (9m) + 6. (9x) – 9 (\(\because\) 10k – 1 is always divisible by 9) 

= 9 (4m + 6x – 1) 

⇒ 10^{k + 1} + 3 (4^{(k + 1) + 2}) + 5 is divisible by 9. 

⇒ T(k + 1) is true whenever T(k) is true, ∀  k∈N 

⇒ 10ⁿ – 3(4^{n + 2}) + 5 is divisible by 9 ∀  k∈N.

Let T (n) = 1 + 2 + 3 + ... + n = \(\frac{1}{2}\) n (n + 1) 

Basic Step: For n = 1, 

LHS = T (1) = 1,   RHS = \(\frac{1}{2}\) × 1 × 2 = 1 ⇒ LHS = RHS ⇒ T(1) is true. 

Induction Step: Assume that T (k) is true, i.e., 

1 + 2 + 3 + ... + k = \(\frac{1}{2}\) k (k + 1) 

To obtain T(k + 1), we add (k + 1)^th term = (k + 1) to both the sides, i.e., 

1 + 2 + 3 + ... + k + (k + 1) = \(\frac{1}{2}\) k (k + 1) + (k + 1) 

⇒ 1 + 2 + 3 + ... + k + (k + 1) = (k + 1) \(\big(\frac{k}{2}+1\big)\)

⇒ 1 + 2 + 3 + ... + k + (k + 1) = \(\frac{1}{2}\) (k + 1) (k + 2) 

⇒ Thus the statement T(n) is true for n = k + 1 under the assumption that T(k) is true. Therefore, by the principle of mathematical induction, the statement is true for every +ve integer n.

Suppose P (n): a + (a + d) + (a + 2d) + … + (a + (n - 1)d) = n/2[2a + (n - 1)d]

Now let us check for n = 1,

P (1): a = 1/2[2a + (1 - 1)d]

: a = a

P (n) is true for n = 1.

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): a + (a + d) + (a + 2d) + … + (a + (k - 1)d) = k/2[2a + (k - 1)d] … (i)

Therefore,

a + (a + d) + (a + 2d) + … + (a + (k - 1)d) + (a + (k)d)

Then, substituting the value of P (k) we get,

= k/2 [2a + (k - 1)d] + (a + kd) by using equation (i)

= [2ka + k(k - 1)d + 2(a + kd)]/2

= [2ka + k²d – kd + 2a + 2kd]/2

= [2ka + 2a + k²d + kd]/2

= [2a(k + 1) + d(k² + k)]/2

= (k + 1)/2 [2a + kd]

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.

Let T(n) be the statement: 

a + ar + ar² + ... + ar^n–1 = \(\frac{a(1-r^n)}{(1-r)}\) , r ≠ 1 

Basic Step: 

For n = 1, ⇒LHS = a, RHS = \(\frac{a(1-r^1)}{(1-r)}\)= a. 

⇒LHS = RHS ⇒ T(1) is true.

Induction Step: 

Let the statement hold true for n = k, i.e., let T (k) be true, i.e., a + ar + ar² + ... + ar^k–1 = \(\frac{a(1-r^k)}{(1-r)}\) 

Then to show T(k + 1) holds, add the (k + 1)^th term = ar^{(k + 1)–1} = ar^k to both the sides of T(k), i.e.,

a + ar + ar² + ... + ar^{k – 1} + ar^k =\(\frac{a(1-r^k)}{(1-r)}\) + ar^k   

\(\frac{a-ar^k+ar^k(1-r)}{(1-r)} = \frac{a-ar^k+ar^k-ar^{k+1}}{(1-r)}\)  =  \(\frac{a-ar^{k+1}}{(1-r)}\) = \(\frac{a(1-r^{k+1})}{(1-r)}\) 

Thus, T(k + 1) is true, whenever T(k) holds true.

Suppose P (n): a + ar + ar² + … + ar^{n – 1} = a[(rⁿ – 1)/(r – 1)]

Now let us check for n = 1,

P (1): a = a (r¹ – 1)/(r - 1)

: a = a

P (n) is true for n = 1.

Then, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): a + ar + ar² + … + ar^{k – 1} = a[(r^k – 1)/(r – 1)] … (i)

Therefore,

a + ar + ar² + … + ar^{k – 1} + ar^k

Then, substituting the value of P (k) we get,

= a[(r^k – 1)/(r – 1)] + ar^k by using equation (i)

= a[r^k – 1 + r^k(r - 1)]/(r-1)

= a[r^k – 1 + r^{k + 1} – r^‑k]/(r - 1)

= a[r^{k + 1} – 1]/(r-1)

P (n) is true for n = k + 1

Thus, P (n) is true for all n ∈ N.