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Question
\[\frac{1}{1 . 6} + \frac{1}{6 . 11} + \frac{1}{11 . 14} + \frac{1}{14 . 19} + . . . + \frac{1}{(5n - 4) (5n + 1)}\]
Solution
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:
\[S_n = \sum^n_{k = 1} \frac{1}{\left( 5k - 4 \right)\left( 5k + 1 \right)}\]
\[ = \frac{1}{5} \sum^n_{k = 1} \left( \frac{1}{\left( 5k - 4 \right)} - \frac{1}{\left( 5k + 1 \right)} \right)\]
\[ = \frac{1}{5} \sum^n_{k = 1} \frac{1}{\left( 5k - 4 \right)} - \frac{1}{5} \sum^n_{k = 1} \frac{1}{\left( 5k + 1 \right)}\]
\[ = \frac{1}{5}\left[ \left( 1 + \frac{1}{6} + \frac{1}{11} + \frac{1}{16} + . . . + \frac{1}{5n - 4} \right) - \left( \frac{1}{6} + \frac{1}{11} + \frac{1}{16} + . . . + \frac{1}{5n - 4} + \frac{1}{5n + 1} \right) \right]\]
\[ = \frac{1}{5}\left[ 1 - \left( \frac{1}{5n + 1} \right) \right]\]
\[ = \frac{n}{5n + 1}\]
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