# Class 11 RD Sharma Solutions – Chapter 12 Mathematical Induction – Exercise 12.2 | Set 2

### Question 17. a + ar + ar2 + … + arn-1  = a [(rn – 1)/(r – 1)], r ≠ 1

Solution:

Let P (n) = a + ar + ar2 + … + arn-1 = a [(rn – 1)/(r – 1)]

Step 1:

Now, let us check P(n) for n = 1.

P (1) = a = a (r1 – 1)/(r – 1)

or, a = a

So, P(1) is true.

Step 2:

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

P (k) = a + ar + ar2 + … + ark-1 = a [(rk – 1)/(r – 1)]     … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

Then, a + ar + ar2 + … + ark-1 + ark

Now, putting the value (i), we get,

= a [(rk – 1)/(r – 1)] + ark              (using eq(i))

= a[rk – 1 + rk(r – 1)] / (r – 1)

= a[rk – 1 + rk+1 – r-k] / (r – 1)

= a[rk+1 – 1] / (r – 1)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 18. a + (a + d) + (a + 2d) + … + (a + (n – 1)d) = n/2 [2a + (n – 1)d]

Solution:

Let P (n) = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) = n/2 [2a + (n – 1)d]

Step 1:

Now, let us check P(n) for n = 1.

P (1) = a = 1/2[2a + (1 – 1)d]

or, a = a

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

Then, a + (a + d) + (a + 2d) + … + (a + (k – 1)d) + (a + (k)d)

Now, putting the value eq(i), we get,

= k/2 [2a + (k – 1)d] + (a + kd)               (Using eq(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]

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 19. 52n – 1 is divisible by 24 for all n ∈ N

Solution:

Let P(n) = 52n – 1 is divisible by 24

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 52 – 1 = 25 – 1 = 24

So, P(1) is true as P(n) is divisible by 24.

Step 2:

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

P (k) = 52k – 1 is divisible by 24

or, 52k – 1 = 24λ   … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e., 52(k+1) – 1 is divisible by 24

Or, 52(k+1) – 1 = 24μ

Then, 52(k + 1) – 1

= 52k.52 – 1

= 25.52k – 1

Now, putting the value eq(i), we get,

= 25.(24λ + 1) – 1           (Using eq(i))

= 25.24λ + 24

= 24λ

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 20. 32n + 7 is divisible by 8 for all n ∈ N

Solution:

Let P (n) =   32n + 7 is divisible by 8

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 32 + 7 = 9 + 7 = 16

So, P(1) is true as P(n) is divisible by 8.

Step 2:

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

P (k) = 32k + 7 is divisible by 8

or, 32k + 7 = 8λ

or, 32k = 8λ – 7   … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

i.e., 32(k+1) + 7 is divisible by 8

or, 32k+2 + 7 = 8μ

Then,

= 32k+2 + 7

= 32k.32 + 7

= 9.32k + 7

Now, putting the value eq(i), we get,

= 9.(8λ – 7) + 7                     (Using eq(i))

= 72λ – 63 + 7

= 72λ – 56

= 8(9λ – 7)

= 8μ

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 21. 52n+2 – 24n – 25 is divisible by 576 for all n ∈ N

Solution:

Let P (n) = 52n+2 – 24n – 25 is divisible by 576

Step 1:

Now, let us check P(n) for n=1.

P (1) = 52.1 + 2 – 24.1 – 25 = 625 – 49 = 576

So, P(1) is true as P (n) is divisible by 576.

Step 2:

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

P (k) = 52k+2 – 24k – 25 is divisible by 576

or, 52k+2 – 24k – 25 = 576λ  … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e., 52k + 4 – 24(k + 1) – 25 is divisible by 576

or, 52k + 2 – 24k – 25 = 576λ

Then,

= 5(2k + 2) + 2 – 24(k + 1) – 25

= 52k+2.52 – 24k – 24 – 25

Now, putting the value eq(i), we get,

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

= 25. 576λ + 576k + 576

= 576(25λ + k + 1)

= 576μ

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 22. 32n + 2 – 8n – 9 is divisible by 8 for all n ∈ N

Solution:

Let P (n) = 32n + 2  – 8n – 9 is divisible by 8

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 32.1 + 2 – 8.1 – 9 = 81 – 17 = 64

So, P(1) is true as P (n) is divisible by 8.

Step 2:

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

P (k) = 32k + 2 – 8k – 9 is divisible by 8

or, 32k + 2 – 8k – 9 = 8λ   … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e., 32k + 4 – 8(k + 1) – 9 is divisible by 8

3(2k + 2) + 2 – 8(k + 1) – 9 = 8μ

Then,

= 32(k+1).32 – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

Now, putting the value (i), we get,

= 72λ + 72k + 81 – 8k – 17              (using eq(i))

= 8(9λ + 8k + 8)

= 8μ

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 23. (ab)n = an bn for all n ∈ N

Solution:

Let P (n) = (ab)n = an bn

Step 1:

Now, let us check P(n) for n = 1.

P (1) = ab = ab

So, P(1) is true.

Step 2:

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

P (k) = (ab)k = ak bk   … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e., (ab)k+1 = ak+1.bk+1

Then,

= (ab)k+1

= (ab)k .(ab)

Now, putting the value eq(i), we get,

= (ak bk) (ab)                (using eq(i))

= (ak+1) (bk+1)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 24. n (n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

Solution:-

Let P (n) = n (n + 1) (n + 5) is a multiple of 3

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 1 (1 + 1) (1 + 5)

or, 2 × 6

or, 12

So, P(1) is true as P(n) is multiple of 3.

Step 2:

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

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

or, k(k + 1) (k + 5) = 3λ   … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

i.e., (k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

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

Then,

(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

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 25. 72n + 23n-3.3n-1 is divisible by 25 for all n ∈ N

Solution:

Let P (n) = 72n + 23n-3 . 3n-1 is divisible by 25

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 72 + 20.30

or, 49 + 1 = 50

So, P(1) is true as P (n) is divisible by 25.

Step 2:

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

P (k) = 72k + 23k-3. 3k-1 is divisible by 25

or, 72k + 23k-3. 3k-1 = 25λ   … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e., 72(k+1) + 23k. 3k is divisible by 25

or, 72k+2 + 23k. 3k = 25μ

Then,

72(k+1) + 23k. 3k

= 72k.72 + 23k. 3k

Now, putting the value (i), we get,

= (25λ – 23k-3. 3k-1) 49 + 23k. 3k                 (Using eq(i))

= 25λ. 49 – 23k/8. 3k/3. 49 + 23k. 3k

= 24×25×49λ – 23k . 3k . 49 + 24 . 23k.3k

= 24×25×49λ – 25 . 23k. 3k

= 25(24 . 49λ – 23k. 3k)

= 25μ

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 26. 2.7n + 3.5n – 5 is divisible by 24 for all n ∈ N

Solution:-

Let P (n) = 2.7n + 3.5n – 5 is divisible by 24

Step 1:

Now, let us check P(n) for n=1.

P (1) = 14 + 15 – 5 = 24

So, P(1) is true as P (n) is divisible by 24.

Step 2:

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

P (k) = 2.7k + 3.5k – 5 is divisible by 24

or, 2.7k + 3.5k – 5 = 24q  … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e., 2.7k+1 + 3.5k+1 – 5

⇒ 2.7.7k + 3.5.5k– 5

Now, we will add and subtract by 3.7.5k- 7.5,

or, 2.7.7k + 3.5.5k – 5 + 3.7.5k – 7.5 – (3.7.5k – 7.5)

or, 7(2.7k + 3.5k – 5) + 3.5.5k – 5 – (3.7.5k – 7.5)

Now from equation (1) we have

or, 7(24q) + 3.5.5k- 5 – 3.7.5k+ 7.5

or, 7(24q) – 2.3.5k- 5 + 35

or, 7(24q) – 2.3.5k+ 30

or, 7(24q) – 6(5k- 5)

Now, we know that (5k – 5) is a multiple of 4.

So we can write (4p) where (p) is from natural number.

or, 7(24q) – 6(4p)

or, 24(7q – p)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 27. 11n+2 + 122n+1 is divisible by 133 for All N ∈ N.

Solution:

Let P (n) = 11n+2+ 122n+1 is Divisible by 133

Step 1:

Now, let us check P(n) for n=1.

P (1) = 111+2 + 122+1= 1331 + 1728 = 3059

So, P(1) is true as P (n) is divisible by 133.

Step 2:

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

P (k) = 11k+2+ 122k+1 is Divisible by 133

or, 11k+2+ 122k+1 = 133p  … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

i.e., 11k+3+ 122k+3

= 11k+2. 11 + 122k+1. 122 + 11. 122k+1 − 11. 122k+1

= 11 (11k+2 + 122k+1) + 122k+1(144 − 11)

Now from equation (i) we have

= 11. 133p + 122k+1. 133               [From eq(i)]

= 133 (11p + 122k+1)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 28. 1×1! + 2×2! + 3×3! +…+ n×n! = (n + 1)! – 1 for all n ∈ N

Solution:

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

Step 1:

Now, let us check P(n) for n=1.

P (1) = 1 × 1! = (2)! – 1 = 1

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(m + 1) is true.

i.e., 1×1! + 2×2! + 3×3! +…+ m×m! + (m + 1)×(m + 1)!

So, = (m + 1)! – 1 + (m + 1) × (m + 1)!

= (m + 1)!(m + 1 + 1) – 1

= (m + 1)!(m + 2) – 1

= (m + 2)! – 1

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 29. Prove that n3 – 7n + 3 is divisible by 3 for all n ∈ N.

Solution:

Let P(n) = n3 – 7n + 3 is divisible by 3

Step 1:

Now, let us check P(n) for n=1.

P(1) = 13 − (7 × 1) + 3 = 1 − 7 + 3 = −3

So, P(1) is true as P(n) is divisible by 3.

Step 2:

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

P (k) = k3 – 7k + 3 is divisible by 3

or, k3 – 7k + 3 = 3m  … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

i.e.,  = (k + 1)3 − 7 (k + 1) + 3

= k3 + 3k2 + 3k + 1 − 7k − 7 + 3

= k3 + 3k2 − 4k − 3

= k3 − 7k + 3 + 3k2 + 3k − 6

Now from equation (i), we have

= 3m + 3 (k2 + k + 2)

= 3 (m + k2 + k + 2)

= 3t              (Here, t is any integer)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 30. Prove that 1 + 2 + 22 + … + 2n = 2n+1 – 1 for all n ∈ N

Solution:

Let P (n) = 1 + 2 + 22 + … + 2n = 2n+1 – 1

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 1 + 2 = 21+1 – 1

or, 3 = 3

So, P(1) is true.

Step 2:

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

P (k) = 1 + 2 + 22 + … + 2k = 2k+1 – 1 … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

i.e., 1 + 2 + 22 + … + 2k+1 = 2k+2 – 1

So, 1 + 2 + 22 + … + 2k + 2k+1

Now from equation (i), we have

or, 2k+1 – 1 + 2k+1

or, 2.2k+1 – 1

or, 2k+2 – 1

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 31. 7 + 77 + 777 + … + 777 . . . . . . . . . . . n − Digits 7 = 7/81 (10 n+1 − 9n − 10)

Solution:

Let P (n) = 7 + 77 + 777 + … + 777 . . . . . . . . . . . n − Digits 7 = 7/81 (10 n+1 − 9n − 10)

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 7 = 7/81 (102 – 9 – 10)

or, 7 = 7/81 ( 100 – 19)

or, 7 = 7/81 (81)

or, 7 = 7

So, P(1) is true.

Step 2:

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

P (k) =  7 + 77 + 777 + … + 777 . . . . . . . . . . . k− Digits7 = 7/81 (10 k+1 − 9k − 10)  … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

i.e., 7 + 77 + 777 + … + 777 . . . . . . . . . . . (k + 1) − Digits7

So, it is basically G.P. (geometric progression) with k + 1 terms.

So, sum P(k + 1)

or, 7/9[9 + 99 + 999 + ……………..+ (k + 1) term]

or, 7/9 [10 – 1 + 100 – 1 +………………..+ (k + 1)term]

or, 7/9 [10 + 100 + 1000 +…………+ (k + 1)term – (1 + 1 + 1 +….+ (k + 1) times)]

or, 7/81 [10k+2 – 9(k+1) – 10]

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 32. Prove that (n7/7) + (n5/5) + (n3/3) + (n2/2) – (37/210)n is a positive integer for all n ∈ N.

Solution:

Let P(n) be (n7/7) + (n5/5) + (n3/3) + (n2/2) – (37/210)n

Step 1:

Now, let us check P(n) for n = 1.

So, P(1) = 1/7+1/5+1/3+1/2-37/210 =1

So, P(1) is true.

Step 2:

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

So, let P(m) be (m7/7) + (m5/5) + (m3/3) + (m2/2) – (37/210)m

Now, (m7/7) + (m5/5) + (m3/3) + (m2/2) – (37/210)m = λ, where λ ∈ N is a positive integer.

Step 3:

Now, we have to prove for P(m+1) is true.

P(m + 1) = ((m + 1)7/7) + ((m + 1)5/5) + ((m + 1)3/3) + ((m + 1)2/2) – (37/210)(m + 1)

= 1/7(m7 + 7m6 + 21m5 + 35m4 + 35m3 + 21m2 + 7m + 1) +

1/5(m5 + 5m4 + 10m3 + 10m2 + 5m + 1) + 1/3(m3 + 3m2 + 3m + 1) +

1/2(m2 + 2m + 1) + 37/210(m + 1)

= (m7/7 + m5/5 + m3/3 + m2/2 − 37/210m) + m6 + 3m5 + 5m4 + 5m3 + 3m2 +

m + m4 + 2m3 + 2m2 + m + m2 + m + m + 1/7 + 1/5 + 1/3 + 1/2 – 37/210

= λ + 3m5 + 5m4 + 5m3 + 3m2 + m + m4 + 2m3 + 2m2 + m + m2 + m + m + 1

As λ is positive, so it is a positive integer.

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

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