Use the formulas $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ and
$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$
of n . $\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$

Name: Use the formulas $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ and $\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$ of n . $\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$
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Question

Use the formulas $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ and
$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$
of  n .  $\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$

Quizplus · Accepted Answer

The Answer of Use the formulas $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ and
$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$
of  n .  $\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$

Use the Formulas ∑k=1nk=n(n+1)2\sum_{k=1}^{n} k=\frac{n(n+1)}{2}∑k=1n​k=2n(n+1)​ And ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}∑k=1n​k2=6n(n+1)(2n+1)​ Of N ∑k=1n(4k2+2k−9)\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)∑k=1n​(4k2+2k−9)

Use the Formulas $\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ And $\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$ Of N $\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$