\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . .\]

#### Solution

Let \[T_n\] be the nth term of the given series.

Thus, we have:

\[T_n = \frac{1}{(3n - 2) (3n + 1)}\]

Now, let

\[S_n\]

be the sum of *n* terms of the given series.

Thus, we have:

\[S_n = \sum^n_{k = 1} \frac{1}{\left( 3k - 2 \right)\left( 3k + 1 \right)}\]

\[ = \frac{1}{3} \sum^n_{k = 1} \left( \frac{1}{\left( 3k - 2 \right)} - \frac{1}{\left( 3k + 1 \right)} \right)\]

\[ = \frac{1}{3} \sum^n_{k = 1} \frac{1}{\left( 3k - 2 \right)} - \frac{1}{3} \sum^n_{k = 1} \frac{1}{\left( 3k + 1 \right)}\]

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

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

\[ = \frac{n}{3n + 1}\]