Use the summation properties to evaluate the series. The following rules may be needed: $$\sum _ { i = 1 } ^ { n } i = \frac { n ( n + 1 ) } { 2 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 3 } = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } .$$
-$$\sum _ { i = 1 } ^ { 3 } \left( - 4 - 2 i ^ { 3 } \right)$$
A) - 150
B) - 84
C) - 216
D) 132

Question

Quizplus · Accepted Answer

The given series can be separated into two parts: $\sum _ { i = 1 } ^ { 3 } ( - 4 )$ and $\sum _ { i = 1 } ^ { 3 } ( - 2 i ^ { 3 } )$. For the first part, $\sum _ { i = 1 } ^ { 3 } ( - 4 ) = - 4 \times 3 = - 12$. For the second part, using the formula for the sum of cubes, $\sum _ { i = 1 } ^ { 3 } i ^ { 3 } = \frac { 3 ^ { 2 } ( 3 + 1 ) ^ { 2 } } { 4 } = \frac { 9 \times 16 } { 4 } = 36$, and then multiplying by -2 gives $-2 \times 36 = -72$. Adding both parts gives $-12 + (-72) = -84$.

Use the Summation Properties to Evaluate the Series - ∑i=13(−4−2i3)\sum _ { i = 1 } ^ { 3 } \left( - 4 - 2 i ^ { 3 } \right)i=1∑3​(−4−2i3)

Use the Summation Properties to Evaluate the Series - $\sum _ { i = 1 } ^ { 3 } \left( - 4 - 2 i ^ { 3 } \right)$