Use the summation properties to evaluate the series. The following rules may be needed: $$\sum _ { i = 1 } ^ { n } i = \frac { n ( n + 1 ) } { 2 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 3 } = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 }$$
-$$\sum _ { i = 1 } ^ { 3 } \left( - 3 i ^ { 2 } + i \right)$$
A) $- 48$
B) $- 36$
C) 20
D) $- 60$

Question

Quizplus · Accepted Answer

To evaluate the series, we apply the given summation properties separately for each term. First, for the $i$ term: $\sum _ { i = 1 } ^ { 3 } i = \frac { 3 ( 3 + 1 ) } { 2 } = \frac { 12 } { 2 } = 6$. Then, for the $i^2$ term: $\sum _ { i = 1 } ^ { 3 } i^2 = \frac { 3 ( 3 + 1 ) ( 2 \cdot 3 + 1 ) } { 6 } = \frac { 3 \cdot 4 \cdot 7 } { 6 } = 14$. The series becomes: $\sum _ { i = 1 } ^ { 3 } ( - 3 i^2 + i ) = -3 \cdot 14 + 6 = -42 + 6 = -36$.

Use the Summation Properties to Evaluate the Series - ∑i=13(−3i2+i)\sum _ { i = 1 } ^ { 3 } \left( - 3 i ^ { 2 } + i \right)i=1∑3​(−3i2+i)

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