Use the summation properties to evaluate the series. The following rules may be needed: $$\sum _ { i = 1 } ^ { n } i = \frac { n ( n + 1 ) } { 2 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 3 } = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } .$$
-$$\sum _ { i = 1 } ^ { 5 } \left( - 4 i ^ { 2 } - i ^ { 3 } \right)$$
A) - 5340
B) - 445
C) - 1780
D) 5

Question

Quizplus · Accepted Answer

To evaluate the series, we apply the given summation properties separately for each term. For $\sum _ { i = 1 } ^ { 5 } -4i^2$, we use the second rule and multiply the result by -4. For $\sum _ { i = 1 } ^ { 5 } -i^3$, we use the third rule. Calculating each and adding them gives the total sum.

Use the Summation Properties to Evaluate the Series - ∑i=15(−4i2−i3)\sum _ { i = 1 } ^ { 5 } \left( - 4 i ^ { 2 } - i ^ { 3 } \right)i=1∑5​(−4i2−i3)

Use the Summation Properties to Evaluate the Series - $\sum _ { i = 1 } ^ { 5 } \left( - 4 i ^ { 2 } - i ^ { 3 } \right)$