Question

How do you find the 4-th partial sum of the infinite series `sum_(n=1)^oo(3/2)^n` ?

Answer 1

`195/16`

Explanation:

When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence

`a_n = (3/2)^n`

Which means that `n`-th term is generates by raising `3/2` to the `n`-th power.

Moreover, the `n`-th partial sum means to sum the first `n` terms from the sequence.

So, in your case, you're looking for `a_1+a_2+a_3+a_4`, which means

`3/2 + (3/2)^2 + (3/2)^3 + (3/2)^4`

You may compute each term, but there is a useful formula:

`sum_{i=1}^n k^i= \frac{k^{n+1}-1}{k-1}`

So, in your case

`sum_{i=0}^4 (3/2)^i= \frac{(3/2)^{5}-1}{3/2-1} = 211/16`

Except you are not including `a_0 = (3/2)^0 = 1` in your sum, so we must subtract it:

`sum_{i=0}^4 (3/2)^i = sum_{i=1}^4 (3/2)^i - 1 = 211/16 - 1 = 195/16`

Answer 2

`sum_1^(n=4)(3/2)^n = 12.1875`

Explanation:

This is geometric progression series of which

first term is `a=3/2=1.5` , common ratio is `r=1.5`

`4` th partial sum ; i.e `n=4`

Sum `S= a * (r^n-1)/(r-1)= 1.5 * ((1.5^4-1)/(1.5-1))=12.1875`

`sum_1^(n=4)(3/2)^n = 12.1875` [Ans]