Use power series operations to find the Taylor series at x = 0 for the given function.
-$$\frac { x ^ { 7 } } { 1 + 2 x }$$
A) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 ^ { n } x _ { x ^ { n } + 7 }$
B) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n _ { 2 } } n _ { x } n + 8$
C) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 ^ { n } n _ { x } + 8$
D) $\sum _ { \mathrm { n } = 0 } ^ { \infty } ( - 1 ) ^ { \mathrm { n } _ { 2 } \mathrm { n } _ { \mathrm { x } } \mathrm { n } + 7 }$

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The Answer of Use power series operations to find the Taylor series at x = 0 for the given function.
-$$\frac { x ^ { 7 } } { 1 + 2 x }$$
A) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 ^ { n } x _ { x ^ { n } + 7 }$
B) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n _ { 2 } } n _ { x } n + 8$
C) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } 2 ^ { n } n _ { x } + 8$
D) $\sum _ { \mathrm { n } = 0 } ^ { \infty } ( - 1 ) ^ { \mathrm { n } _ { 2 } \mathrm { n } _ { \mathrm { x } } \mathrm { n } + 7 }$

Use Power Series Operations to Find the Taylor Series at X