How do you find the sum of $4 {(\frac{3}{2})}^{i - 1}$ $4(3/2)^(i-1)$ from i=0 to 6?

Question

How do you find the sum of $4 {(\frac{3}{2})}^{i - 1}$ $4(3/2)^(i-1)$ from i=0 to 6?

1 Answer

Impact of this question

1757 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-18T08:23:30

$\frac{2059}{24}$ $2059/24$

Explanation:

This is a geometric series $n \sum i = 0 a r^{i}$ $sum_{i=0}^n ar^i$ with first term

$a = 4 {(\frac{3}{2})}^{- 1} = \frac{8}{3}$ $a = 4(3/2)^{-1}=8/3$

and common ratio $r = \frac{3}{2}$ $r = 3/2$

The sum is

$u m_{i = 0}^{n} a r^{i} = a \frac{r^{n + 1} - 1}{r - 1} = \frac{8}{3} \frac{{(\frac{3}{2})}^{7} - 1}{\frac{3}{2} - 1} = \frac{8}{3} \frac{\frac{2187}{128} - 1}{\frac{1}{2}} = \frac{16}{3 \times 128} (2187 - 128) = \frac{2059}{24}$ $um_{i=0}^n ar^i = a{r^{n+1}-1}/{r-1} = 8/3 {(3/2)^7-1}/{3/2-1} = 8/3 {2187/128-1}/{1/2} = 16/{3\times 128}(2187-128)= 2059/24$

Science

Math

Humanities

... and beyond

How do you find the sum of $4 {(\frac{3}{2})}^{i - 1}$ $4(3/2)^(i-1)$ from i=0 to 6?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the sum of 4(3/2)^(i-1)4(32)i−14(3/2)^(i-1) from i=0 to 6?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the sum of $4 {(\frac{3}{2})}^{i - 1}$ $4(3/2)^(i-1)$ from i=0 to 6?