Question

For all n≥1, prove that 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/n(n+1) = n/n+1​

Answer 1

Answer

We can write as,

$\sf \dashrightarrow P(n) : \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} +...+ \dfrac{1}{n(n + 1)} = \dfrac{n}{n + 1}$

$\sf \dashrightarrow P(1) : \dfrac{1}{1.2} = \dfrac{1}{2.3} = \dfrac{1}{3.4}, which \: is \: true$

Thus, P(n) is true for n = 1.

Assume that P(k) is true for some natural number k, i.e,

$\sf \dashrightarrow \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} +...+ \dfrac{1}{k(k + 1)} = \dfrac{k}{k + 1 \: - - - - (i)}$

Add (k+1)th term on both sides, we get

$\sf \dashrightarrow \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} +...+ \dfrac{1}{k(k + 1)} = \dfrac{1}{(k + 1) (k + 2)}$

$\sf \dashrightarrow \bigg[ \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} +...+ \dfrac{1}{k(k + 1)} \bigg] + \dfrac{1}{(k + 1)(k + 2)}$

$\sf \dashrightarrow \dfrac{k}{k + 1} + \dfrac{1}{(k + 1)(k + 2)}$

$\sf \dashrightarrow \dfrac{k(k + 2) + 1}{(k + 1) (k + 2)}$

$\sf \dashrightarrow \dfrac{({k}^{2} + 2k + 1)}{(k + 1)(k + 2)}$

$\sf \dashrightarrow \dfrac{(k + 1)^{2}}{(k + 1)(k + 2)}$

$\sf \dashrightarrow \dfrac{(k + 1)}{(k + 2)}$

$\sf \dashrightarrow \dfrac{(k + 1)}{(k + 1) + 1}$

$\sf \dashrightarrow \dfrac{n}{n + 1} 1M$

Thus is true. Hence, by the principle of mathematical induction, P(n) is true for all natural numbers.

Answer 2

Step-by-step explanation:

We can write this as,

[1/1 - 1/2] + [1/2 - 1/3] + [1/3 - 1/4] ... [1/n+ 1/(n+1)]

By simplifing we get

1/1 - 1/(n+1)

=> [(n+1) -1]/(n+1)

=> n/n+1

Hence Proved.