Use the summation properties to evaluate the series. The following rules may be needed: $$\sum _ { i = 1 } ^ { n } i = \frac { n ( n + 1 ) } { 2 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 3 } = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } .$$
-$$\sum _ { i = 1 } ^ { 6 } \left( i ^ { 2 } + i - 4 \right)$$
A) 417
B) 108
C) 88
D) 437

Question

Quizplus · Accepted Answer

The given series can be broken down into three separate summations using the properties of summation: $\sum _ { i = 1 } ^ { 6 } i ^ { 2 } + \sum _ { i = 1 } ^ { 6 } i - \sum _ { i = 1 } ^ { 6 } 4$. Using the provided formulas and evaluating each part gives: $\frac{6(6+1)(2*6+1)}{6} + \frac{6(6+1)}{2} - 4*6 = 91 + 21 - 24 = 88$.

Use the Summation Properties to Evaluate the Series - ∑i=16(i2+i−4)\sum _ { i = 1 } ^ { 6 } \left( i ^ { 2 } + i - 4 \right)i=1∑6​(i2+i−4)

Use the Summation Properties to Evaluate the Series - $\sum _ { i = 1 } ^ { 6 } \left( i ^ { 2 } + i - 4 \right)$