Solve the following

Question:

$\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.14}+\frac{1}{14.19}+\ldots+\frac{1}{(5 n-4)(5 n+1)}$

Solution:

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=\frac{1}{(5 n-4)(5 n+1)}$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} \frac{1}{(5 k-4)(5 k+1)}$

$=\frac{1}{5} \sum_{k=1}^{n}\left(\frac{1}{(5 k-4)}-\frac{1}{(5 k+1)}\right)$

$=\frac{1}{5} \sum_{k=1}^{n} \frac{1}{(5 k-4)}-\frac{1}{5} \sum_{k=1}^{n} \frac{1}{(5 k+1)}$

$=\frac{1}{5}\left[\left(1+\frac{1}{6}+\frac{1}{11}+\frac{1}{16}+\ldots+\frac{1}{5 n-4}\right)-\left(\frac{1}{6}+\frac{1}{11}+\frac{1}{16}+\ldots+\frac{1}{5 n-4}+\frac{1}{5 n+1}\right)\right]$

$=\frac{1}{5}\left[1-\left(\frac{1}{5 n+1}\right)\right]$

$=\frac{n}{5 n+1}$

 

 

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