Use the formulas $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$, and $\sum_{i=1}^{n}
-\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$
A) 14,355
B) 5280
C) 11,625
D) 4125

Question

Use the formulas $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$, and $\sum_{i=1}^{n} 
-\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$
A) 14,355
B) 5280
C) 11,625
D) 4125

Quizplus · Accepted Answer

The Answer of Use the formulas $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$, and $\sum_{i=1}^{n} 
-\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$
A) 14,355
B) 5280
C) 11,625
D) 4125

Use the Formulas ∑i=1ni=n(n+1)2,∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}∑i=1n​i=2n(n+1)​,∑i=1n​i2=6n(n+1)(2n+1)​ , And \sum_{i=1}^{n} -\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)

Use the Formulas $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$ , And $\sum_{i=1}^{n} -\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$