Find the sum using the formulas for the sums of powers of integers. $\sum _ { n = 1 } ^ { 11 } \left( 8 n - 4 n ^ { 2 } \right)$
A)-1100
B)-4488
C)-396
D)264
E)-1496

Question

Quizplus · Accepted Answer

The given sum can be simplified and solved using the formulas for the sums of powers of integers. The sum is $\sum _ { n = 1 } ^ { 11 } \left( 8 n - 4 n ^ { 2 } \right)$, which can be split into two separate sums: $8\sum _ { n = 1 } ^ { 11 } n - 4\sum _ { n = 1 } ^ { 11 } n^2$.The formula for the sum of the first $m$ positive integers is $\frac{m(m+1)}{2}$, and the formula for the sum of the squares of the first $m$ positive integers is $\frac{m(m+1)(2m+1)}{6}$.Applying these formulas:- For the first sum: $8\sum _ { n = 1 } ^ { 11 } n = 8 \cdot \frac{11 \cdot 12}{2} = 8 \cdot 66 = 528$- For the second sum: $4\sum _ { n = 1 } ^ { 11 } n^2 = 4 \cdot \frac{11 \cdot 12 \cdot 23}{6} = 4 \cdot 506 = 2024$The total sum is $528 - 2024 = -1496$.

Find the Sum Using the Formulas for the Sums of Powers