Notes: `sum_(i = 1)^12 16(1/2)^(i - 1)` Find the sum of the finite geometric series.

Thursday, July 2, 2015

The given geometric series is:

$\displaystyle{\sum_{{{i}={1}}}^{{12}}}{16}{{\left(\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}$

Since the summation notation is in the form

$\displaystyle{\sum_{{{i}={1}}}^{{i}}}{a}_{{1}}\cdot{{r}}^{{{i}-{1}}}$

it can be seen that the first term and common ratio of the geometric series is a1=16 and r=1/2.

Applying the summation formula of finite geometric series

$\displaystyle{S}_{{i}}={a}_{{1}}\cdot\frac{{{1}-{{r}}^{{i}}}}{{{1}-{r}}}$

the sum of the first twelve terms of this series is:

$\displaystyle{S}_{{12}}={16}\cdot\frac{{{1}-{{\left(\frac{{1}}{{2}}\right)}}^{{12}}}}{{{1}-\frac{{1}}{{2}}}}=\frac{{4095}}{{128}}$

Therefore, $\displaystyle{\sum_{{{i}={1}}}^{{12}}}{16}{{\left(\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}=\frac{{4095}}{{128}}$ .

Notes