Solve the following

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

Thus, we have:

$T_{n}=\frac{1}{(3 n-2)(3 n+1)}$

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

Thus, we have:

$S_{n}=\sum_{k=1}^{n} \frac{1}{(3 k-2)(3 k+1)}$

$=\frac{1}{3} \sum_{k=1}^{n}\left(\frac{1}{(3 k-2)}-\frac{1}{(3 k+1)}\right)$

$=\frac{1}{3} \sum_{k=1}^{n} \frac{1}{(3 k-2)}-\frac{1}{3} \sum_{k=1}^{n} \frac{1}{(3 k+1)}$

$=\frac{1}{3}\left[\left(1+\frac{1}{4}+\frac{1}{7}+\frac{1}{10}+\ldots+\frac{1}{3 n-2}\right)-\left(\frac{1}{4}+\frac{1}{7}+\frac{1}{10}+\ldots+\frac{1}{3 n-2}+\frac{1}{3 n+1}\right)\right]$

$=\frac{1}{3}\left[1-\left(\frac{1}{3 n+1}\right)\right]$

$=\frac{n}{3 n+1}$