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$\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$

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$\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$

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

Essay

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge

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)

Essay

Correct Answer:VerifiedUnlock this answer nowGet Access to more Verified Answers free of chargeAccess For Free

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 $\sum_{k=1}^{n}\left(4 k^{2}+2 k-9\right)$

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge