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Find the sum of $n$ terms of the series whose $n^{th}$ term is given by $n^2+2^n$.

$\begin{array}{1 1}\large\frac{n(n+1)(2n+1)}{6}+2.(2^n-1) \\ \large\frac{n(n+1)(2n+1)}{6}+2.(2^{n-1}-1) \\ \large\frac{n(n+1)(2n+1)}{6}+.(2^{n+1}-1) \\\large\frac{n(n+1)(2n+1)}{6}+.(2^n-1)\end{array} $

1 Answer

  • Sum of $n$ terms of any series is $=S_n=\sum\:t_n$
  • $\sum (A+B)=\sum A+\sum B$
  • $\sum n^2=1^2+2^2+3^2+..........=\large\frac{n(n+1)(2n+1)}{6}$
  • Sum of $n$ terms of a G.P$=\large\frac{a.(r^n-1)}{r-1}$
Given $t_n=n^2+2^n$.
$\therefore\:$ Sum of $n$ terms of the series $=S_n=\sum t_n=\sum (n^2+2^n)$
We know that $\sum (A+B)=\sum A+\sum B$
$=\sum n^2+\sum 2^n$
$\sum n^2=1^2+2^2+3^2+..........=\large\frac{n(n+1)(2n+1)}{6}$
$\sum 2^n=2+2^2+2^3+2^4+..........$ which is a G.P. with common ratio $2$
answered Mar 27, 2014 by rvidyagovindarajan_1

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