Finding sums of infinite series as stated in the title the series are infinite and im struggle with how to find the common ratio. and first term (kinda) well I want it to be on the form: ar^k where r is the ratio and a the first term
(3+2^n)/(2^(n+2)) (n=1)
any tips on how to get that series on the form ar^k ?

It may help to decompose the series:
$$S = \displaystyle \sum_{n = 1}^{k} \frac{3 + 2^n}{2^{n + 2}} = \displaystyle \sum_{n = 1}^{k} \frac{3}{2^{n + 2}} + \displaystyle \sum_{n = 1}^{k} \frac{1}{4}$$
Going a bit further,
$$S = (\frac{3}{8} + \frac{3}{16} + \frac{3}{32} + \dots + \frac{3}{2^{k+2}}) + (\frac{1}{4} k)$$
So in this case, $a = \frac{3}{8}$ and $r = \frac{1}{2}$. Additionally, the infinite series diverges using the $n$-th term test ($\lim_{n \to \infty} \frac{1}{4} = \frac{1}{4} \neq 0$)
