Question

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

Answer 1

Let $T_n$ be the nth term of the given series.

Thus, we have:

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

Now, let $S_n$ be the sum of n terms of the given series.

Thus, we have:
$$\begin{gathered} S_n=\sum _{k=1}^n\frac{1}{\left(5k-4\right)\left(5k+1\right)} \\ =\frac{1}{5}\sum _{k=1}^n\left(\frac{1}{\left(5k-4\right)}-\frac{1}{\left(5k+1\right)}\right) \\ =\frac{1}{5}\sum _{k=1}^n\frac{1}{\left(5k-4\right)}-\frac{1}{5}\sum _{k=1}^n\frac{1}{\left(5k+1\right)} \\ =\frac{1}{5}\left[\left(1+\frac{1}{6}+\frac{1}{11}+\frac{1}{16}+...+\frac{1}{5n-4}\right)-\left(\frac{1}{6}+\frac{1}{11}+\frac{1}{16}+...+\frac{1}{5n-4}+\frac{1}{5n+1}\right)\right] \\ =\frac{1}{5}\left[1-\left(\frac{1}{5n+1}\right)\right] \\ =\frac{n}{5n+1} \end{gathered}$$