# Partial Sums of Infinite Series

## Key Questions

• #### Answer:

$\frac{195}{16}$

#### Explanation:

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

${a}_{n} = {\left(\frac{3}{2}\right)}^{n}$

Which means that $n$-th term is generates by raising $\frac{3}{2}$ 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

$\frac{3}{2} + {\left(\frac{3}{2}\right)}^{2} + {\left(\frac{3}{2}\right)}^{3} + {\left(\frac{3}{2}\right)}^{4}$

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

${\sum}_{i = 1}^{n} {k}^{i} = \setminus \frac{{k}^{n + 1} - 1}{k - 1}$

So, in your case

${\sum}_{i = 0}^{4} {\left(\frac{3}{2}\right)}^{i} = \setminus \frac{{\left(\frac{3}{2}\right)}^{5} - 1}{\frac{3}{2} - 1} = \frac{211}{16}$

Except you are not including ${a}_{0} = {\left(\frac{3}{2}\right)}^{0} = 1$ in your sum, so we must subtract it:

${\sum}_{i = 0}^{4} {\left(\frac{3}{2}\right)}^{i} = {\sum}_{i = 1}^{4} {\left(\frac{3}{2}\right)}^{i} - 1 = \frac{211}{16} - 1 = \frac{195}{16}$

• A partial sum ${S}_{n}$ of an infinite series ${\sum}_{i = 1}^{\infty} {a}_{i}$ is the sum of the first n terms, that is,
${S}_{n} = {a}_{1} + {a}_{2} + {a}_{3} + \cdots + {a}_{n} = {\sum}_{i = 1}^{n} {a}_{i}$.