Use the summation properties to evaluate the series. The following rules may be needed: $$\begin{array} { l l l }
\mathrm { n } & \mathrm { n } & \mathrm { n } \
\sum \mathrm { i } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 } ; & \sum \mathrm { i } ^ { 2 } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 } ; & \sum \mathrm { i } ^ { 3 } = \frac { \mathrm { n } ^ { 2 } ( \mathrm { n } + 1 ) ^ { 2 } } { 4 } \
\mathrm { i } = 1 & \mathrm { i } = 1 & \mathrm { i } = 1
\end{array}$$
-$$\sum _ { i = 1 } ^ { 6 } \left( - 3 i ^ { 2 } - i ^ { 3 } \right)$$
A) $- 714$
B) 168
C) $- 2856$
D) $- 8568$

Question

Use the summation properties to evaluate the series. The following rules may be needed: $$\begin{array} { l l l } 
\mathrm { n } & \mathrm { n } & \mathrm { n } \
\sum \mathrm { i } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 } ; & \sum \mathrm { i } ^ { 2 } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 } ; & \sum \mathrm { i } ^ { 3 } = \frac { \mathrm { n } ^ { 2 } ( \mathrm { n } + 1 ) ^ { 2 } } { 4 } \
\mathrm { i } = 1 & \mathrm { i } = 1 & \mathrm { i } = 1
\end{array}$$
-$$\sum _ { i = 1 } ^ { 6 } \left( - 3 i ^ { 2 } - i ^ { 3 } \right)$$
A) $- 714$
B) 168
C) $- 2856$
D) $- 8568$

Quizplus · Accepted Answer

To evaluate the series $\sum _ { i = 1 } ^ { 6 } \left( - 3 i ^ { 2 } - i ^ { 3 } \right)$, we use the given summation properties. First, we separate the series into two parts:$$\sum _ { i = 1 } ^ { 6 } \left( - 3 i ^ { 2 } \right) + \sum _ { i = 1 } ^ { 6 } \left( - i ^ { 3 } \right)$$Using the given formulas:$$\sum _ { i = 1 } ^ { 6 } i ^ { 2 } = \frac { 6 ( 6 + 1 ) ( 2 \cdot 6 + 1 ) } { 6 } = \frac { 6 \cdot 7 \cdot 13 } { 6 } = 7 \cdot 13 = 91$$$$\sum _ { i = 1 } ^ { 6 } i ^ { 3 } = \frac { 6 ^ { 2 } ( 6 + 1 ) ^ { 2 } } { 4 } = \frac { 36 \cdot 49 } { 4 } = 9 \cdot 49 = 441$$Then, we multiply by the coefficients and add the results:$$-3 \cdot 91 - 441 = -273 - 441 = -714$$Therefore, the correct answer is $-714$.

Use the Summation Properties to Evaluate the Series - ∑i=16(−3i2−i3)\sum _ { i = 1 } ^ { 6 } \left( - 3 i ^ { 2 } - i ^ { 3 } \right)i=1∑6​(−3i2−i3)

Use the Summation Properties to Evaluate the Series - $\sum _ { i = 1 } ^ { 6 } \left( - 3 i ^ { 2 } - i ^ { 3 } \right)$