Use the summation properties to evaluate the series. The following rules may be needed: $$\begin{array} { l l l }
\mathrm { n } & \mathrm { n } & \mathrm { n } \
\sum \mathrm { i } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 } ; & \sum \mathrm { i } ^ { 2 } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 } ; & \sum \mathrm { i } ^ { 3 } = \frac { \mathrm { n } ^ { 2 } ( \mathrm { n } + 1 ) ^ { 2 } } { 4 } \
\mathrm { i } = 1 & \mathrm { i } = 1 & \mathrm { i } = 1
\end{array}$$
-$$\sum _ { i = 1 } ^ { 4 } \left( i + i ^ { 2 } \right)$$
A) 110
B) 400
C) 100
D) 40

Question

Use the summation properties to evaluate the series. The following rules may be needed: $$\begin{array} { l l l } 
\mathrm { n } & \mathrm { n } & \mathrm { n } \
\sum \mathrm { i } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 } ; & \sum \mathrm { i } ^ { 2 } = \frac { \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { 6 } ; & \sum \mathrm { i } ^ { 3 } = \frac { \mathrm { n } ^ { 2 } ( \mathrm { n } + 1 ) ^ { 2 } } { 4 } \
\mathrm { i } = 1 & \mathrm { i } = 1 & \mathrm { i } = 1
\end{array}$$
-$$\sum _ { i = 1 } ^ { 4 } \left( i + i ^ { 2 } \right)$$
A) 110
B) 400
C) 100
D) 40

Quizplus · Accepted Answer

The given series can be split into two separate summations: $\sum _ { i = 1 } ^ { 4 } i + \sum _ { i = 1 } ^ { 4 } i ^ { 2 }$. Using the provided formulas, $\sum _ { i = 1 } ^ { 4 } i = \frac { 4 ( 4 + 1 ) } { 2 } = 10$ and $\sum _ { i = 1 } ^ { 4 } i ^ { 2 } = \frac { 4 ( 4 + 1 ) ( 2 \cdot 4 + 1 ) } { 6 } = 30$. Adding these results gives $10 + 30 = 40$.

Use the Summation Properties to Evaluate the Series - ∑i=14(i+i2)\sum _ { i = 1 } ^ { 4 } \left( i + i ^ { 2 } \right)i=1∑4​(i+i2)

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