$$\dfrac{1}{3.7} +\dfrac{ 1}{7.11} + \dfrac{1}{11.15} + … + \dfrac{1}{(4n-1)(4n+3)} = \dfrac{n}{3(4n+3)}$$

Asked by Aaryan | 1 year ago |  80

##### Solution :-

Let us check for n = 1,

P (1): $$\dfrac{ 1}{3.7} = \dfrac{1}{(4.1-1)(4+3)}$$

$$\dfrac{ 1}{21} = \dfrac{1}{21}$$

P (n) is true for n =1.

Now, let us check for P (n) is true for n = k, and have to prove that P (k + 1) is true.

P (k): $$\dfrac{ 1}{3.7} + \dfrac{1}{7.11 }+ \dfrac{1}{11.15} + … + \dfrac{1}{(4k-1)(4k+3)} = \dfrac{k}{3(4k+3)}$$ …. (i)

Substituting the value of P (k) we get,

=$$\dfrac{ k}{(4k+3) }+ \dfrac{1}{(4k+3)(4k+7)}$$

= $$\dfrac{\dfrac{1}{(4k+3) k(4k+7)+3}}{3(4k+7)}$$

=$$\dfrac{ (k+1) }{ 3(4k+7)}$$

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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