Simplifying $(3/2)^i$ for $i = 0$ to $\\(log_2 n) -1$

Question

$$\sum_{i=0}^{\\(log_2 n)-1} \frac{3^i}{2^i}$$

I have been working on solving a recurrence, and finally simplified to this step. However, I am lost as to where to proceed now. I know the sum of powers of 2, but I don't know what to do since it's in the denominator.

Welcome to Mathematics Stack Exchange. Do you know the sum of powers of $\dfrac3 2$? — J. W. Tanner, Jun 24 '19 at 20:40
Have I edited the title appropriately? Please change it if not, Ardvark. — The Count, Jun 24 '19 at 20:57
@J.W.Tanner no I don't, isn't that just the answer? How do you find the sum of powers of a fraction — Ardvark243, Jun 25 '19 at 15:43
that's the answer; do you know how to sum powers of $5$? $a$? — J. W. Tanner, Jun 25 '19 at 15:48

score 2 · Answer 1 · answered Jun 24 '19 at 20:40

2

Hint $$\sum_{i=0}^{N}r^i=\frac{1-r^{N+1}}{1-r}$$ when $r\ne 1$

answered Jun 24 '19 at 20:40

DINEDINE

6,081

score 0 · Answer 2 · answered Jun 25 '19 at 15:46

0

$$\sum_{i=0}^{\log_2 n} \frac{3^i}{2^i}=\sum_{i=0}^{\log_2 n} \left(\frac{3}{2}\right)^i.$$

Now use the formula for the sum of a geometric series.

answered Jun 25 '19 at 15:46

J. W. Tanner

60,406

Simplifying $(3/2)^i$ for $i = 0$ to $\\(log_2 n) -1$

2 Answers2