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

### Prove the following by the principle of mathematical induction:

### Question 17. a + ar + ar^{2} + … + ar^{n-1 } = a [(r^{n} – 1)/(r – 1)], r ≠ 1

**Solution:**

Let P (n) = a + ar + ar

^{2}+ … + ar^{n-1 }= a [(r^{n}– 1)/(r – 1)]Step 1:

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

P (1) = a = a (r

^{1 }– 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 + … + ar

^{k-1}= a [(r^{k}– 1)/(r – 1)] … (i)Step 3:

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

Then, a + ar + ar

^{2}+ … + ar^{k-1}+ ar^{k}Now, putting the value (i), we get,

= a [(r

^{k}– 1)/(r – 1)] + ar^{k}(using eq(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)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. 5^{2n} – 1 is divisible by 24 for all n ∈ N

**Solution:**

Let P(n) = 5

^{2n}– 1 is divisible by 24Step 1:

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

P (1) = 5

^{2}– 1 = 25 – 1 = 24So, 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) = 5

^{2k }– 1 is divisible by 24or, 5

^{2k}– 1 = 24λ … (i)Step 3:

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

i.e., 5

^{2(k+1) }– 1 is divisible by 24Or, 5

^{2(k+1) }– 1 = 24μThen, 5

^{2(k + 1) }– 1= 5

^{2k}.5^{2}– 1= 25.5

^{2k}– 1Now, 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. 3^{2n} + 7 is divisible by 8 for all n ∈ N

**Solution:**

Let P (n) = 3

^{2n}+ 7 is divisible by 8Step 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) = 3

^{2k}+ 7 is divisible by 8or, 3

^{2k}+ 7 = 8λor, 3

^{2k}= 8λ – 7 … (i)Step 3:

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

i.e., 3

^{2(k+1) }+ 7 is divisible by 8or, 3

^{2k+2 }+ 7 = 8μThen,

= 3

^{2k+2}+ 7= 3

^{2k}.3^{2}+ 7= 9.3

^{2k}+ 7Now, 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. 5^{2n+2} – 24n – 25 is divisible by 576 for all n ∈ N

**Solution:**

Let P (n) = 5

^{2n+2}– 24n – 25 is divisible by 576Step 1:

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

P (1) = 5

^{2.1 + 2}– 24.1 – 25 = 625 – 49 = 576So, 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) = 5

^{2k+2}– 24k – 25 is divisible by 576or, 5

^{2k+2}– 24k – 25 = 576λ … (i)Step 3:

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

i.e., 5

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

^{2k + 2 }– 24k – 25 = 576λThen,

= 5

^{(2k + 2) + 2 }– 24(k + 1) – 25= 5

^{2k+2}.5^{2}– 24k – 24 – 25Now, 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. 3^{2n + 2 }– 8n – 9 is divisible by 8 for all n ∈ N

**Solution:**

Let P (n) = 3

^{2n + 2 }– 8n – 9 is divisible by 8Step 1:

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

P (1) = 3

^{2.1 + 2 }– 8.1 – 9 = 81 – 17 = 64So, 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) = 3

^{2k + 2}– 8k – 9 is divisible by 8or, 3

^{2k + 2}– 8k – 9 = 8λ … (i)Step 3:

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

i.e., 3

^{2k + 4}– 8(k + 1) – 9 is divisible by 83

^{(2k + 2) + 2}– 8(k + 1) – 9 = 8μThen,

= 3

^{2(k+1)}.3^{2}– 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} = a^{n} b^{n} for all n ∈ N

**Solution:**

Let P (n) = (ab)

^{n}= a^{n}b^{n}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}= a^{k}b^{k}… (i)Step 3:

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

i.e., (ab)

^{k+1}= a^{k+1}.b^{k+1}Then,

= (ab)

^{k+1}= (ab)

^{k }.(ab)Now, putting the value eq(i), we get,

= (a

^{k}b^{k}) (ab) (using eq(i))= (a

^{k+1}) (b^{k+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. 7^{2n} + 2^{3n-3}.3^{n-1} is divisible by 25 for all n ∈ N

**Solution:**

Let P (n) = 7

^{2n}+ 2^{3n-3}. 3^{n-1}is divisible by 25Step 1:

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

P (1) = 7

^{2}+ 2^{0}.3^{0}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) = 7

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

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

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

i.e., 7

^{2(k+1)}+ 2^{3k}. 3^{k}is divisible by 25or, 7

^{2k+2}+ 2^{3k}. 3^{k}= 25μThen,

7

^{2(k+1)}+ 2^{3k}. 3^{k}= 7

^{2k}.7^{2}+ 2^{3k}. 3^{k}Now, putting the value (i), we get,

= (25λ – 2

^{3k-3}. 3^{k-1}) 49 + 2^{3k}. 3^{k}(Using eq(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

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. 2.7^{n} + 3.5^{n} – 5 is divisible by 24 for all n ∈ N

**Solution:-**

Let P (n) = 2.7

^{n}+ 3.5^{n}– 5 is divisible by 24Step 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) = 2.7

^{k}+ 3.5^{k}– 5 is divisible by 24or, 2.7

^{k}+ 3.5^{k}– 5 = 24q … (i)Step 3:

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

i.e., 2.7

^{k+1}+ 3.5^{k+1}– 5⇒ 2.7.7

^{k }+ 3.5.5^{k}– 5Now, we will add and subtract by 3.7.5k- 7.5,

or, 2.7.7

^{k }+ 3.5.5^{k }– 5 + 3.7.5^{k }– 7.5 – (3.7.5^{k }– 7.5)or, 7(2.7

^{k }+ 3.5^{k }– 5) + 3.5.5^{k }– 5 – (3.7.5^{k }– 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. 11^{n+2} + 12^{2n+1} is divisible by 133 for All N ∈ N.

**Solution:**

Let P (n) = 11

^{n+2}+ 12^{2n+1}is Divisible by 133Step 1:

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

P (1) = 11

^{1+2}+ 12^{2+1}= 1331 + 1728 = 3059So, 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) = 11

^{k+2}+ 12^{2k+1}is Divisible by 133or, 11

^{k+2}+ 12^{2k+1 }= 133p … (i)Step 3:

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

i.e., 11

^{k+3}+ 12^{2k+3}= 11

^{k+2}. 11 + 12^{2k+1}. 12^{2}+ 11. 12^{2k+1}− 11. 12^{2k+1}= 11 (11

^{k+2}+ 12^{2k+1}) + 12^{2k+1}(144 − 11)Now from equation (i) we have

= 11. 133p + 12

^{2k+1}. 133 [From eq(i)]= 133 (11p + 12

^{2k+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 n^{3} – 7n + 3 is divisible by 3 for all n ∈ N.

**Solution:**

Let P(n) = n

^{3}– 7n + 3 is divisible by 3Step 1:

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

P(1) = 1

^{3}− (7 × 1) + 3 = 1 − 7 + 3 = −3So, 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) = k

^{3}– 7k + 3 is divisible by 3or, k

^{3}– 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= k

^{3}+ 3k^{2}+ 3k + 1 − 7k − 7 + 3= k

^{3}+ 3k^{2}− 4k − 3= k

^{3}− 7k + 3 + 3k^{2}+ 3k − 6Now from equation (i), we have

= 3m + 3 (k

^{2}+ k + 2)= 3 (m + k

^{2}+ 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 + 2^{2} + … + 2^{n} = 2^{n+1} – 1 for all n ∈ N

**Solution:**

Let P (n) = 1 + 2 + 2

^{2}+ … + 2^{n}= 2^{n+1 }– 1Step 1:

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

P (1) = 1 + 2 = 2

^{1+1 }– 1or, 3 = 3

So, P(1) is true.

Step 2:

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

P (k) = 1 + 2 + 2

^{2}+ … + 2^{k}= 2^{k+1}– 1 … (i)Step 3:

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

i.e., 1 + 2 + 2

^{2}+ … + 2^{k+1}= 2^{k+2}– 1So, 1 + 2 + 2

^{2}+ … + 2^{k }+ 2^{k+1}Now from equation (i), we have

or, 2

^{k+1 }– 1 + 2^{k+1 }or, 2.2

^{k+1 }– 1or, 2

^{k+2 }– 1So, 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 (10

^{2}– 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 [10

^{k+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 (n^{7}/7) + (n^{5}/5) + (n^{3}/3) + (n^{2}/2) – (37/210)n is a positive integer for all n ∈ N.

**Solution:**

Let P(n) be (n

^{7}/7) + (n^{5}/5) + (n^{3}/3) + (n^{2}/2) – (37/210)nStep 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 (m

^{7}/7) + (m^{5}/5) + (m^{3}/3) + (m^{2}/2) – (37/210)mNow, (m

^{7}/7) + (m^{5}/5) + (m^{3}/3) + (m^{2}/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(m

^{7 }+ 7m^{6 }+ 21m^{5 }+ 35m^{4 }+ 35m^{3 }+ 21m^{2 }+ 7m + 1) +1/5(m

^{5 }+ 5m^{4 }+ 10m^{3 }+ 10m^{2 }+ 5m + 1) + 1/3(m^{3 }+ 3m^{2 }+ 3m + 1) +1/2(m

^{2 }+ 2m + 1) + 37/210(m + 1)= (m

^{7}/7 + m^{5}/5 + m^{3}/3 + m^{2}/2 − 37/210m) + m^{6 }+ 3m^{5 }+ 5m^{4 }+ 5m^{3 }+ 3m^{2 }+m + m

^{4 }+ 2m^{3 }+ 2m^{2 }+ m + m^{2 }+ m + m + 1/7 + 1/5 + 1/3 + 1/2 – 37/210= λ + 3m

^{5 }+ 5m^{4 }+ 5m^{3 }+ 3m^{2 }+ m + m^{4 }+ 2m^{3 }+ 2m^{2 }+ m + m^{2 }+ m + m + 1As 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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