1. using 1/(1 - r)=∑(n = 0)^∞ r^n, a power - series representation of 1/(1 - t^12) is ∑(n = 0)^∞(t^12)^n. 2…

1. using 1/(1 - r)=∑(n = 0)^∞ r^n, a power - series representation of 1/(1 - t^12) is ∑(n = 0)^∞(t^12)^n. 2. since 1/(1 - t^12)=∑(n = 0)^∞( ), a power - series representation of t/(1 - t^12)=t(1/(1 - t^12)) is ∑(n = 0)^∞( ). submit skip (you cannot come back) i help? read it submit answer

1. using 1/(1 - r)=∑(n = 0)^∞ r^n, a power - series representation of 1/(1 - t^12) is ∑(n = 0)^∞(t^12)^n. 2. since 1/(1 - t^12)=∑(n = 0)^∞( ), a power - series representation of t/(1 - t^12)=t(1/(1 - t^12)) is ∑(n = 0)^∞( ). submit skip (you cannot come back) i help? read it submit answer

Answer

Explanation:

Step1: Recall geometric - series formula

The geometric - series formula is $\frac{1}{1 - r}=\sum_{n = 0}^{\infty}r^{n}$, for $|r|\lt1$. When $r = t^{12}$, we have $\frac{1}{1 - t^{12}}=\sum_{n = 0}^{\infty}(t^{12})^{n}=\sum_{n = 0}^{\infty}t^{12n}$.

Step2: Find power - series for $\frac{t}{1 - t^{12}}$

Since $\frac{t}{1 - t^{12}}=t\times\frac{1}{1 - t^{12}}$, and $\frac{1}{1 - t^{12}}=\sum_{n = 0}^{\infty}t^{12n}$, then $\frac{t}{1 - t^{12}}=t\sum_{n = 0}^{\infty}t^{12n}=\sum_{n = 0}^{\infty}t^{12n + 1}$.

Answer:

For the first blank in the first part: $t^{12}$; for the first blank in the second part: $t^{12n}$; for the second blank in the second part: $t^{12n+1}$