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By the principle of mathematical induction, prove1.2 + 2.3 + 3.4 + … + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N. |
Answer» Let P(n) denote the statement 1.2 + 2.3 + 3.4 + …… + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\) Put n = 1 LHS = 1(1 + 1) = 2 RHS = \(\frac{1(1+1)(1+2)}{3}\) = \(\frac{1(2)(3)}{3}\) = 2 ∴ P(1) is true. Now assume that the statement be true for n = k (i.e.,) assume P(k) be true (i.e.,) assume 1.2 + 2.3 + 3.4 + … + k(k + 1) = \(\frac{k(k+1)(k+2)}{3}\) be true To prove: P(k + 1) is true (i.e.,) to prove: 1.2 + 2.3 + 3.4 + … + k(k + 1) + (k + 1) (k + 2) = \(\frac{(k+1)(k+2)(k+3)}{3}\) Consider 1.2 + 2.3 + 3.4 + ... + k(k + 1) + (k + 1) (k + 2) = [1.2 + 23 + … + k(k + 1)] + (k + 1) (k + 2) = \(\frac{k(k+1)(k+2)}{3}\) + (k + 1) (k + 2) = \(\frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\) = \(\frac{(k+1)(k+2)(k+3)}{3}\) ∴ P(k + 1) is true. Thus if P(k) is true, P(k + 1) is true. By the principle of Mathematical ‘induction, P(n) is true for all n ∈ N. 1.2 + 2.3 + 3.4 + … + n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\) |
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