How to get a closed form for $\sum_{i=1}^n \frac{3^i}{2^i}$

Question

How can I get a closed form for:

$$\sum_{i=1}^n \frac{3^i}{2^i}$$

I have just started studying closed forms for summations and I am still lost on this matter so I would appreciate if you guys could explain that step by step or suggest me some references for me to study.

Thanks in advance.

Answer 1

Hint

You have $$\sum_{i=1}^n \frac{3^i}{2^i}=\sum_{i=1}^n \left(\dfrac 32\right)^i$$ and (if $r\ne 1$) $$\sum_{i=1}^n r^n=\dfrac{r^{n+1}-r}{r-1}.$$ Thus ... (please, complete).

Answer 2

You can write $\frac{a^i}{b^i}$ as $(\frac{a}{b})^i$

thus

$$\sum_{i=1}^n \frac{3^i}{2^i}=\sum_{i=1}^n \left(\dfrac 32\right)^i$$ $$\sum_{i=1}^n \frac{3^i}{2^i}=\dfrac 32+\left(\dfrac 32\right)^2+\left(\dfrac 32\right)^3+\cdots+\left(\dfrac 32\right)^n$$

this is geometric series with $r=\frac{3}{2}$

$$S=\dfrac 32+\left(\dfrac 32\right)^2+\left(\dfrac 32\right)^3+\cdots+\left(\dfrac 32\right)^n$$ multiply both sides by $\frac{3}{2}$

$$\frac{3}{2}S=\hspace{12pt}\left(\dfrac 32\right)^2+\left(\dfrac 32\right)^3+\cdots+\left(\dfrac 32\right)^n+\left(\dfrac 32\right)^{n+1}$$

subtract both equations

$$\frac{3}{2}S-S=\left(\dfrac 32\right)^{n+1}-\dfrac 32$$ $$\frac{S}{2}=\frac{3}{2}\left(\left(\dfrac 32\right)^{n}-1\right)$$ $$S=3\left(\left(\dfrac 32\right)^{n}-1\right)$$