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Question
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . .\]
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:
\[S_n = \sum^n_{k = 1} \frac{1}{\left( 3k - 2 \right)\left( 3k + 1 \right)}\]
\[ = \frac{1}{3} \sum^n_{k = 1} \left( \frac{1}{\left( 3k - 2 \right)} - \frac{1}{\left( 3k + 1 \right)} \right)\]
\[ = \frac{1}{3} \sum^n_{k = 1} \frac{1}{\left( 3k - 2 \right)} - \frac{1}{3} \sum^n_{k = 1} \frac{1}{\left( 3k + 1 \right)}\]
\[ = \frac{1}{3}\left[ \left( 1 + \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + . . . + \frac{1}{3n - 2} \right) - \left( \frac{1}{4} + \frac{1}{7} + \frac{1}{10} + . . . + \frac{1}{3n - 2} + \frac{1}{3n + 1} \right) \right]\]
\[ = \frac{1}{3}\left[ 1 - \left( \frac{1}{3n + 1} \right) \right]\]
\[ = \frac{n}{3n + 1}\]
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