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Sequences and Series (p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Sequences and Series (p-1)2 Marks
Question
Find $\sum_{r=1}^n\left(5 r^2+4 r-3\right)$.
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Answer

$\begin{aligned} & \sum_{r=1}^n\left(5 r^2+4 r-3\right) \\ = & 5 \sum_{r=1}^n r^2+4 \sum_{r=1}^n r-3 \sum_{r=1}^n 1 \\ = & 5 \cdot \frac{n(n+1)(2 n+1)}{6}+4 \cdot \frac{n(n+1)}{2}-3 n \\ = & \frac{n}{6}\left[5\left(2 n^2+3 n+1\right)+12(n+1)-18\right] \\ = & \frac{n}{6}\left(10 n^2+15 n+5+12 n+12-18\right) \\ = & \frac{n}{6}\left(10 n^2+27 n-1\right)\end{aligned}$

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