$$\dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + … + \dfrac{1}{n(n+1)} = \dfrac{n}{(n+1)}$$

Asked by Aaryan | 1 year ago |  72

##### Solution :-

For, n = 1

P (n) = $$\dfrac{1}{1.2} = \dfrac{1}{1+1}$$

$$\dfrac{1}{2}= \dfrac{1}{2}$$

P (n) is true for n = 1

Let’s check for P (n) is true for n = k,

$$\dfrac{ 1}{1.2} + \dfrac{1}{2.3} +\dfrac{ 1}{3.4 }+ … + \dfrac{1}{k(k+1)} + \dfrac{k}{(k+1) (k+2)} = \dfrac{(k+1)}{(k+2)}$$

= $$\dfrac{ \dfrac{1}{(k+1)}}{(k+2)} + \dfrac{k}{(k+1)}$$

=$$\dfrac{ (k+1) }{ (k+2)}$$

P (n) is true for n = k + 1

Hence, P (n) is true for all n ∈ N.

Answered by Aaryan | 1 year ago

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