The sum 1+3+11+25+45+71+ upto 20 terms, is equal to

Question

The sum $1+3+11+25+45+71+\ldots$ upto 20 terms, is equal to

Accepted Answer

$$\begin{aligned}
&T_r=3 r^2-7 r+5 \text { using second order difference }\
&\begin{aligned}
& \sum_{r=1}^{20}\left(3 r^2-7 r+5\right)=3 \Sigma r^2-7 \Sigma r+5 \Sigma(1) \
& =\frac{3(n)(n+1)(2 n+1)}{6}-\frac{7 n(n+1)}{2}-5 n, n=20 \
& =7240
\end{aligned}
\end{aligned}$$

Solutions · 1