MFC UNIT-1--1. Mathematical Induction

Mathematical Induction
1.	The sum of the series 1³ + 2³ + 3³ + ………..n³ is (a) {(n + 1)/2}² (b) {n/2}² (c) n(n + 1)/2 (d) {n(n + 1)/2}² Answer Answer: (d) {n(n + 1)/2}² Hint: Given, series is 1³ + 2³ + 3³ + ……….. n³ Sum = {n(n + 1)/2}²	D
2.	If n is an odd positive integer, then an + bn is divisible by : (a) a² + b² (b) a + b (c) a – b (d) none of these Answer: (b) a + b Hint: Given number = an + bn Let n = 1, 3, 5, …….. an + bn = a + b an + bn = a³ + b³ = (a + b) × (a² + b² + ab) and so on. Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,….. So, the given number is divisible by (a + b)	B
3.	1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)} (a) n(n + 1) (b) n/(n + 1) (c) 2n/(n + 1) (d) 3n/(n + 1) Answer Answer: (b) n/(n + 1) Hint: Let the given statement be P(n). Then, P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)} = n/(n + 1). Putting n = 1 in the given statement, we get LHS = 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2. LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} = k/(k + 1) ..…(i) Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)} [1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)} = k/(k + 1)+1/{ (k + 1)(k + 2)}. {k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii) = {k(k + 2) + 1}/{(k + 1)(k + 2} = {(k + 1)² }/{(k + 1)(k + 2)} = (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1) ⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)} = (k + 1)/(k + 1 + 1) ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.	B
4.	The sum of the series 1² + 2² + 3² + ………..n² is (a) n(n + 1)(2n + 1) (b) n(n + 1)(2n + 1)/2 (c) n(n + 1)(2n + 1)/3 (d) n(n + 1)(2n + 1)/6 Answer Answer: (d) n(n + 1)(2n + 1)/6 Hint: Given, series is 1² + 2² + 3² + ………..n² Sum = n(n + 1)(2n + 1)/6	D
5.	{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = (a) 1/(n + 1) for all n ∈ N. (b) 1/(n + 1) for all n ∈ R (c) n/(n + 1) for all n ∈ N. (d) n/(n + 1) for all n ∈ R Answer Answer: (a) 1/(n + 1) for all n ∈ N. Hint: Let the given statement be P(n). Then, P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1). When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½. Therefore LHS = RHS. Thus, P(1) is true. Let P(k) be true. Then, P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1) Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}] = [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}] = [1/(k + 1)] ∙ [(k + 1)/(k + 2)] = 1/(k + 2) Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2) ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.	A
6.	For any natural number n, 7n – 2n is divisible by (a) 3 (b) 4 (c) 5 (d) 7 Answer Answer: (c) 5 Hint: Given, 7n – 2n Let n = 1 7n – 2n = 71 – 21 = 7 – 2 = 5 which is divisible by 5 Let n = 2 7n – 2n = 72 – 22 = 49 – 4 = 45 which is divisible by 5 Let n = 3 7n – 2n = 73 – 23 = 343 – 8 = 335 which is divisible by 5 Hence, for any natural number n, 7n – 2n is divisible by 5	C
7.	1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} = (a) {n(n + 3)}/{4(n + 1)(n + 2)} (b) (n + 3)/{4(n + 1)(n + 2)} (c) n/{4(n + 1)(n + 2)} (d) None of these Answer Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)} Hint: Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} . Putting n = 1 in the given statement, we get LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6. Therefore LHS = RHS. Thus, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. ……. (i) Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)} = [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1)(k + 2}] + 1/{(k + 1)(k + 2)(k + 3)} = [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)] = {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)} = (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)} = {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)} = {(k + 1)(k + 4)}/{4(k + 2)(k + 3) ⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)} = {(k + 1)(k + 2)}/{4(k + 2)(k + 3)} ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.	A
8.	The nth terms of the series 3 + 7 + 13 + 21 +………. is (a) 4n – 1 (b) n² + n + 1 (c) none of these (d) n + 2 Answer Answer: (b) n² + n + 1 Hint: Let S = 3 + 7 + 13 + 21 +……….an-1 + an …………1 and S = 3 + 7 + 13 + 21 +……….an-1 + an …………2 Subtract equation 1 and 2, we get S – S = 3 + (7 + 13 + 21 +……….an-1 + an) – (3 + 7 + 13 + 21 +……….an-1 + an) ⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (an – an-1) – an ⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – an ⇒ an = 3 + {4 + 6 + 8 + ……(n-1)terms} ⇒ an = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2} ⇒ an = 3 + (n – 1)/2 × {8 + (n – 2)2} ⇒ an = 3 + (n – 1) × {4 + n – 2} ⇒ an = 3 + (n – 1) × (n + 2) ⇒ an = 3 + n² + n – 2 ⇒ an = n² + n + 1 So, the nth term is n² + n + 1	B
9.	n(n + 1)(n + 5) is a multiple of ____ for all n ∈ N (a) 2 (b) 3 (c) 5 (d) 7 Answer Answer: (b) 3 Hint: Let P(n) : n(n + 1)(n + 5) is a multiple of 3. For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3. So, the given statement is true for n = 1, i.e. P(1) is true. Let P(k) be true. Then, P(k) : k(k + 1)(k + 5) is a multiple of 3 ⇒ K(k + 1)(k + 5) = 3m for some natural number m, … (i) Now, (k + 1)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2) = k(k + 1)(k + 2) + 6(k + 1)(k + 2) = k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2) = k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2) = k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification] = 3m + 3(k + 1 )(k + 4) [using (i)] = 3[m + (k + 1)(k + 4)], which is a multiple of 3 ⇒ P(k + 1) : (k + 1 )(k + 2)(k + 6) is a multiple of 3 ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.	B
10.	(n² + n) is ____ for all n ∈ N. (a) Even (b) odd (c) Either even or odd (d) None of these Answer Answer: (a) Even Hint: Let P(n): (n² + n) is even. For n = 1, the given expression becomes (1² + 1) = 2, which is even. So, the given statement is true for n = 1, i.e., P(1)is true. Let P(k) be true. Then, P(k): (k² + k) is even ⇒ (k² + k) = 2m for some natural number m. ….. (i) Now, (k + 1)² + (k + 1) = k² + 3k + 2 = (k² + k) + 2(k + 1) = 2m + 2(k + 1) [using (i)] = 2[m + (k + 1)], which is clearly even. Therefore, P(k + 1): (k + 1)² + (k + 1) is even ⇒ P(k + 1) is true, whenever P(k) is true. Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n)is true for all n ∈ N	A