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Use the Binomial Series to Find the Maclaurin Series for the Function

Question 58

Multiple Choice

Use the binomial series to find the Maclaurin series for the function
f(x) =8(1+x) 2f ( x ) = \frac { 8 } { ( 1 + x ) ^ { 2 } }


A) 8(x+1) 2=8+8n=0135(2n1) xn2nn!\frac { 8 } { ( x + 1 ) ^ { 2 } } = 8 + 8 \sum _ { n = 0 } ^ { \infty } \frac { 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) x ^ { n } } { 2 ^ { n } n ! }
B) 8(x+1) 2=8+8n=1135(2n1) xn2nn!\frac { 8 } { ( x + 1 ) ^ { 2 } } = 8 + 8 \sum _ { n = 1 } ^ { \infty } \frac { 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) x ^ { n } } { 2 ^ { n } n ! }
C) 8(x+1) 2=8+8n=0135(2n1) xnn!\frac { 8 } { ( x + 1 ) ^ { 2 } } = 8 + 8 \sum _ { n = 0 } ^ { \infty } \frac { 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) x ^ { n } } { n ! }
D) 8(x+1) 2=8n=0135(2n1) xn2nn!\frac { 8 } { ( x + 1 ) ^ { 2 } } = 8 \sum _ { n = 0 } ^ { \infty } \frac { 1 \cdot 3 \cdot 5 \cdots ( 2 n - 1 ) x ^ { n } } { 2 ^ { n } n ! }
E) 8(x+1) 2=8n=1(2n1) xn2n\frac { 8 } { ( x + 1 ) ^ { 2 } } = 8 \sum _ { n = 1 } ^ { \infty } \frac { ( 2 n - 1 ) x ^ { n } } { 2 ^ { n } }

Correct Answer:

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