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

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

Use Power Series Operations to Find the Taylor Series at X