How do you find the fourth partial sum of Sigma 5(1/2)^i from i=1 to oo?

Jul 12, 2018

$\frac{75}{16}$

Explanation:

A series is an infinite sum, i.e. a sum with infinite terms.

Partial sums are approximations of the series, where you reduce it to a finite number of terms. In fact, the ${n}^{\text{th}}$ partial sum is the sum of the first $n$ terms, and the series is the limit of the partial sums as $n \setminus \to \setminus \infty$. You can imagine this, for example, as approaching the infinite sum by summing the first $10$, $100$, $1000$, $100000$, $1000000000$,... terms.

So, the fourth partial sum is the sum of the first $4$ terms. Since $i$ starts from one, we must sum the terms corresponding to $i = 1 , 2 , 3 , 4$.

First of all, let's factor the $5$ out of the sum:

$\setminus {\sum}_{i = 1}^{4} 5 {\left(\frac{1}{2}\right)}^{i} = 5 \setminus {\sum}_{i = 1}^{4} {\left(\frac{1}{2}\right)}^{i}$

Now we can simply translate the notations, plugging $1 , 2 , 3 , 4$ in place of $i$ for each term:

$5 \setminus {\sum}_{i = 1}^{4} {\left(\frac{1}{2}\right)}^{i} = 5 \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}\right) 5 \cdot \frac{15}{16} = \frac{75}{16}$