Use the summation properties to evaluate the series. The following rules may be needed: $$\sum _ { i = 1 } ^ { n } i = \frac { n ( n + 1 ) } { 2 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } ; \quad \sum _ { i = 1 } ^ { n } i ^ { 3 } = \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 }$$
-$$\sum _ { i = 1 } ^ { 4 } \left( 4 i ^ { 2 } + 3 i - 1 \right)$$
A) 146
B) 149
C) 154
D) 126

Question

Quizplus · Accepted Answer

To evaluate the series, split it into separate sums: $\sum_{i=1}^{4} (4i^2 + 3i - 1) = 4\sum_{i=1}^{4} i^2 + 3\sum_{i=1}^{4} i - \sum_{i=1}^{4} 1$. Using the given formulas: $\sum_{i=1}^{4} i^2 = \frac{4(4+1)(2*4+1)}{6} = 30$, $\sum_{i=1}^{4} i = \frac{4(4+1)}{2} = 10$, and $\sum_{i=1}^{4} 1 = 4$. Thus, the series evaluates to $4*30 + 3*10 - 4 = 120 + 30 - 4 = 146$.

Use the Summation Properties to Evaluate the Series - ∑i=14(4i2+3i−1)\sum _ { i = 1 } ^ { 4 } \left( 4 i ^ { 2 } + 3 i - 1 \right)i=1∑4​(4i2+3i−1)

Use the Summation Properties to Evaluate the Series - $\sum _ { i = 1 } ^ { 4 } \left( 4 i ^ { 2 } + 3 i - 1 \right)$