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_{\mathrm{i}=1}^{15}\left(\mathrm{i}^{2}+\mathrm{i}\right)$
A) 14,520
B) 18,120
C) 5440
D) 1360

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_{\mathrm{i}=1}^{15}\left(\mathrm{i}^{2}+\mathrm{i}\right)$
A) 14,520
B) 18,120
C) 5440
D) 1360

Quizplus · Accepted Answer

The given sum can be split into two separate sums: $\sum_{i=1}^{15} i^2 + \sum_{i=1}^{15} i$. Using the formulas, we find $\frac{15(15+1)(2*15+1)}{6} + \frac{15(15+1)}{2}$, which simplifies to $1240 + 120 = 1360$.

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_{\mathrm{i}=1}^{15}\left(\mathrm{i}^{2}+\mathrm{i}\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_{\mathrm{i}=1}^{15}\left(\mathrm{i}^{2}+\mathrm{i}\right)$