Notes On Application of Mathematical Induction - CBSE Class 11 Maths
Principle of Mathematical Induction Let P(n) be a mathematical statement, where n is a natural number, such that: (i) The statement is true for n = 1, or P(1) is true. (ii) If the statement is true for n = k, where k is a positive integer, then the statement is also true for n = k + 1. P(k) is true ⇒ P(k + 1) is true. Then, P(n) is true for all natural numbers n. Principle one is a statement of fact, while principle two is a condition. If P(n) is true for all n ≥ 2, then step 1 starts from n = 2 and we verify the result of P(2). Next, if the second principle is true for n = k, then it is also true for n = k + 1. Using the principle of mathematical induction, prove the following statements: (i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1) (ii) For all n ∈ N; (23n - 1) is divisible by 7. (i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1) P(n): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1) For n = 1, P(1): 1/1.3 = 1/((2x1)+1) ⇒ 1/3 = 1/3 ∴ P(1) is true. Assume that P(n) is true for some k ∈ N. P(k): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) = k/(2k + 1) ----------- (1) To prove that P(k + 1) is true, add the next term of the series to both sides of the statement. 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) + 1/([2(k+1)-1][2(k+1)+1])= k/(2k + 1) + 1/([2(k+1)-1][2(k+1)+1]) = k/(2k + 1)  +  1/((2k + 1)(2k + 3)) = (k/(2k + 3) + 1)/((2k+1)(2k+3)) =(2k2+3k+1)/((2k+1)(2k+3)) =((2k+1)(k+1))/((2k+1)(2k+3)) =(k+1)/(2k+3) =(k+1)/(2(k+1)+1) =P(k+1) ∴ P(k+1) is true. Hence, P(n) is true for all natural numbers n. We have to prove that the expression is divisible by seven for all natural numbers n. P(n): (23n - 1) is divisible by 7. For n = 1, P(1): 23x1 - 1 = 7, which is divisible by 7. ∴ P(1) is true. Let P(n) be true for some k ∈ N. ∴ P(k):(23k -1) is divisible by 7 ------- (1) Need to prove that P(k + 1) is true. P(k + 1): 23(k+1) - 1 = 23k+3 - 1 =23k.23 - 1 =23k.8 - 1 =23k.(7+1) - 1 =23k.7 + (23k - 1) 23k.7is a multiple of 7. 23k-1 is divisible by 7 from equation 1. ∴ P(k+1) is divisible by 7. Hence, (23n - 1) is divisible by 7 for all natural numbers n.

#### Summary

Principle of Mathematical Induction Let P(n) be a mathematical statement, where n is a natural number, such that: (i) The statement is true for n = 1, or P(1) is true. (ii) If the statement is true for n = k, where k is a positive integer, then the statement is also true for n = k + 1. P(k) is true ⇒ P(k + 1) is true. Then, P(n) is true for all natural numbers n. Principle one is a statement of fact, while principle two is a condition. If P(n) is true for all n ≥ 2, then step 1 starts from n = 2 and we verify the result of P(2). Next, if the second principle is true for n = k, then it is also true for n = k + 1. Using the principle of mathematical induction, prove the following statements: (i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1) (ii) For all n ∈ N; (23n - 1) is divisible by 7. (i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1) P(n): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1) For n = 1, P(1): 1/1.3 = 1/((2x1)+1) ⇒ 1/3 = 1/3 ∴ P(1) is true. Assume that P(n) is true for some k ∈ N. P(k): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) = k/(2k + 1) ----------- (1) To prove that P(k + 1) is true, add the next term of the series to both sides of the statement. 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) + 1/([2(k+1)-1][2(k+1)+1])= k/(2k + 1) + 1/([2(k+1)-1][2(k+1)+1]) = k/(2k + 1)  +  1/((2k + 1)(2k + 3)) = (k/(2k + 3) + 1)/((2k+1)(2k+3)) =(2k2+3k+1)/((2k+1)(2k+3)) =((2k+1)(k+1))/((2k+1)(2k+3)) =(k+1)/(2k+3) =(k+1)/(2(k+1)+1) =P(k+1) ∴ P(k+1) is true. Hence, P(n) is true for all natural numbers n. We have to prove that the expression is divisible by seven for all natural numbers n. P(n): (23n - 1) is divisible by 7. For n = 1, P(1): 23x1 - 1 = 7, which is divisible by 7. ∴ P(1) is true. Let P(n) be true for some k ∈ N. ∴ P(k):(23k -1) is divisible by 7 ------- (1) Need to prove that P(k + 1) is true. P(k + 1): 23(k+1) - 1 = 23k+3 - 1 =23k.23 - 1 =23k.8 - 1 =23k.(7+1) - 1 =23k.7 + (23k - 1) 23k.7is a multiple of 7. 23k-1 is divisible by 7 from equation 1. ∴ P(k+1) is divisible by 7. Hence, (23n - 1) is divisible by 7 for all natural numbers n.

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