The sum $1+3+11+25+45+71+\ldots$ upto 20 terms, is equal to
Solution
<p>$$\begin{aligned}
&T_r=3 r^2-7 r+5 \text { using second order difference }\\
&\begin{aligned}
& \sum_{r=1}^{20}\left(3 r^2-7 r+5\right)=3 \Sigma r^2-7 \Sigma r+5 \Sigma(1) \\
& =\frac{3(n)(n+1)(2 n+1)}{6}-\frac{7 n(n+1)}{2}-5 n, n=20 \\
& =7240
\end{aligned}
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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