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  4. The sum of 20 terms of the series 1+(1+3)+(1+3+5)+(1+3+5+7)+ .... is

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Q. The sum of $20$ terms of the series $1+\left(1+3\right)+\left(1+3+5\right)+\left(1+3+5+7\right)+ ....$ is

Sequences and Series

Solution:

$T_{n} = 1+3+5+....+\left(2n-1\right)= n^{2}$
$\therefore S = \sum\limits_{k=1}^{n} T_{k} $
$= \sum \limits_{k=1}^{n} k^{2}$
$ = \frac{n\left(n+1\right)\left(2n+1\right)}{6}$
Put $n= 20$, we get
$ S_{20} = \frac{20\left(21\right)\left(41\right)}{6}$
$ = 2870 $