1.

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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