$$\dfrac{ 1}{2.5} + \dfrac{1}{5.8} + \dfrac{1}{8.11} + … + \dfrac{1}{(3n-1) (3n+2) }=\dfrac{ n}{(6n+4)}$$

Asked by Aaryan | 1 year ago |  123

##### Solution :-

Let us check P (n) is true for n = 1

P (1): $$\dfrac{ 1}{2.5} = \dfrac{1}{6.1}+4$$$$\dfrac {1}{10} =\dfrac {1}{10}$$

P (1) is true.

Now,

Let us check for P (k) is true, and have to prove that P (k + 1) is true.

P (k): $$\dfrac{1}{2.5} +\dfrac{ 1}{5.8} +\dfrac{ 1}{8.11 }+ … +\dfrac{ 1}{(3k-1) (3k+2)} = \dfrac{k}{(6k+4)}$$

$$\dfrac{ k}{(6k+4)} + \dfrac{1}{(3k+2)(3k+5)}$$

$$\dfrac{k(3k+5)+2}{ 2(3k+2)(3k+5)}$$

$$\dfrac{ (k+1) }{ (6(k+1)+4)}$$

P (k + 1) is true.

Hence proved by mathematical induction.

Answered by Aaryan | 1 year ago

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