# How do you find the sum of the finite geometric sequence of sum_(j=1)^6 32(1/4)^(j-1) ?

Jan 1, 2018

Use a formula found in this reference

sum_(j=1)^n ar^(j-1)=a(1-r^n)/(1-r); r!=1

#### Explanation:

Substituting into the formula, $n = 6 , a = 32 \mathmr{and} r = \frac{1}{4}$:

${\sum}_{j = 1}^{6} 32 {\left(\frac{1}{4}\right)}^{j - 1} = 32 \frac{1 - {\left(\frac{1}{4}\right)}^{6}}{1 - \frac{1}{4}}$

${\sum}_{j = 1}^{6} 32 {\left(\frac{1}{4}\right)}^{j - 1} = 32 \frac{1 - \frac{1}{4096}}{1 - \frac{1}{4}}$

${\sum}_{j = 1}^{6} 32 {\left(\frac{1}{4}\right)}^{j - 1} = \frac{1365}{32}$