# How do you find the sum of the finite geometric sequence of sum_(j=1)^12 16(1/2)^(j-1)?

Jul 13, 2017

Given: ${\sum}_{j = 1}^{12} 16 {\left(\frac{1}{2}\right)}^{j - 1}$

This can be written as:

$16 + {\sum}_{j = 1}^{11} 16 {\left(\frac{1}{2}\right)}^{j} \text{ [1]}$

From this reference we obtain the formula:

${S}_{n} = {\sum}_{j = 1}^{n} {a}_{j} = {a}_{1} \frac{1 - {r}^{n}}{1 - r}$

The formula fits the second term of equation [1] where ${a}_{1} = 16 , r = \frac{1}{2} \mathmr{and} n = 11$

${S}_{12} = 16 + 16 \frac{1 - {\left(\frac{1}{2}\right)}^{11}}{1 - \left(\frac{1}{2}\right)}$

${S}_{12} = 47.984375$