Sum of the series $(i)$ and $(ii)$

Question

Find the sum of the series:
$(i)1+\frac{2}{9}+\frac{2.5}{9.18}+\frac{2.5.8}{9.18.27}+\cdots +\infty$
$(ii)1+\frac{3}{4}+\frac{7}{16}+\frac{13}{64}+\cdots +\infty$
The answer providing my book is :$(i)\frac{9}{4}^{\frac{1}{3}},(ii)\frac{8}{3}$.
i couldn't think of how to start.Because i didn't find any helpful pattern to go further.Any hints or solution will be appreciated. Thanks in advance.

Answer 1

I still don't know what it's supposed to be .

1) I supposed that numerator always increased by next even number : $3 = 1 + \color{blue}{2}$, then $7 = 3 +\color{blue}{4}$ and so on. And denominator is $4^{k}$. So we have $\displaystyle \sum_{k=0}^{\infty} \frac{k(k+1)+1}{4^{k}} = \sum_{k=0}^{\infty} \frac{k(k+1)}{4^{k}} + \sum_{k=0}^{\infty} \frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $\displaystyle S(x) = \sum_{k=0}^{\infty} \frac{x^{k+1}}{4^{k}} = x^{1}\sum_{k = 0}^{\infty} \frac{x^{k}}{4^{k}} = \frac{4x^{1}}{4-x}$ , so you have $\displaystyle \sum_{k=0}^{\infty} \frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (\frac{4x}{4-x})''|_{x=1} = \frac{32}{27}$