$\text { Write an equivalent series of the series } \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 6 n + 1 } } { 6 n + 1 } \text { with the index of }$ summation beginning at $n = 5$.
A) $\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 6 n + 29 } } { 6 n + 29 }$
B) $\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 6 n - 31 } } { 6 n - 31 }$
C) $\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n - 5 } x ^ { 6 n - 29 } } { 6 n - 29 }$
D)$\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n - 5 } x ^ { 6 n + 29 } } { 6 n + 29 }$
E)$\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n + 5 } x ^ { 7 n - 29 } } { 7 n - 29 }$

Question

Quizplus · Accepted Answer

To shift the index from $n = 0$ to $n = 5$, substitute $n = m + 5$ in the original series. This gives $\sum_{m=5}^{\infty} \frac{(-1)^{m-5} x^{6(m-5)+1}}{6(m-5)+1}$, which simplifies to $\sum_{n=5}^{\infty} \frac{(-1)^{n-5} x^{6n-29}}{6n-29}$.

Summation Beginning At n=5n = 5n=5 A) ∑n=5∞(−1)nx6n+296n+29\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 6 n + 29 } } { 6 n + 29 }∑n=5∞​6n+29(−1)nx6n+29​

Summation Beginning At $n = 5$
A) $\sum _ { n = 5 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 6 n + 29 } } { 6 n + 29 }$