Question

Prove by mathematical induction
$\frac{1}{1.2} + \frac{1}{2.3} + .... + \frac{1}{3(n + 1)} = \frac{n}{n + 1}$

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Answer 1

Given Sequence,

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

Let check the validity of the sequence for n = 1.

LHS : 1/1 × 2 = 1/2

RHS : 1/( 1 + 1) = 1/2

Therefore, P(1) is true.

Say the sequence is valid for k, where k belongs to natural numbers.

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

To prove that P(k + 1) is true, where k + 1 is a natural number.

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

LHS :

$\sf \bigg( \dfrac{1}{1.2} + \dfrac{1}{2.3} + .... \bigg)+ \dfrac{1}{(k + 1)(k + 2)} \\ \\ \longrightarrow \sf \: \dfrac{k}{k + 1} + \dfrac{1}{(k + 1)(k + 2)} \\ \\ \longrightarrow \sf \: \dfrac{1}{k + 1} \bigg(k + \dfrac{1}{k + 2} \bigg) \\ \\ \longrightarrow \sf \: \dfrac{(k + 1) {}^{2} }{(k + 1)(k + 2)} \\ \\ \longrightarrow \sf \dfrac{k + 1}{k + 2} \longrightarrow \: RHS$

By Principle of Mathematical Induction, P(n) is true for all values of natural numbers.

Answer 2

$\huge\underline\mathbb\color{blue}{Refer\: Attachment}$