Use the summation properties to evaluate the series. The following rules may be needed: $$\sum _ { i = 1 } ^ { n } i = \frac { n ( n + 1 ) } { 2 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 3 } = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 }$$
-$$\sum _ { i = 1 } ^ { 6 } \left( i + i ^ { 2 } \right)$$
A) 462
B) 112
C) 441
D) 1764

Question

Quizplus · Accepted Answer

The given series can be separated into two sums: $\sum _ { i = 1 } ^ { 6 } i + \sum _ { i = 1 } ^ { 6 } i ^ { 2 }$. Using the provided formulas, we find $\sum _ { i = 1 } ^ { 6 } i = \frac { 6 ( 6 + 1 ) } { 2 } = 21$ and $\sum _ { i = 1 } ^ { 6 } i ^ { 2 } = \frac { 6 ( 6 + 1 ) ( 2 \cdot 6 + 1 ) } { 6 } = 91$. Adding these results gives $21 + 91 = 112$.

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

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