Notes: `sum_(i = 1)^7 64(-1/2)^(i - 1)` Find the sum of the finite geometric series.

Sunday, May 25, 2014

$\displaystyle{\sum_{{{i}={1}}}^{{7}}}{64}{{\left(-\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}$ Find the sum of the finite geometric series.

$\displaystyle{\sum_{{{i}={1}}}^{{7}}}{64}{{\left(-\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}$

The given summation notation has a form

$\displaystyle{\sum_{{{i}={1}}}^{{i}}}{a}_{{1}}{{\left({r}\right)}}^{{{i}-{1}}}$

Base on this, the first term and common ratio of the geometric series can be identified. The values are a1=64 and r=-1/2.

Plugging in the values of the a1 and r to the formula of geometric series

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

the sum of the first seven terms will be:

$\displaystyle{S}_{{7}}={64}\cdot\frac{{{1}-{{\left(-\frac{{1}}{{2}}\right)}}^{{7}}}}{{{1}-{\left(-\frac{{1}}{{2}}\right)}}}$

$\displaystyle{S}_{{7}}={64}\cdot{\left(\frac{{{1}-{\left(-\frac{{1}}{{128}}\right)}}}{{{1}-{\left(-\frac{{1}}{{2}}\right)}}}\right.}$

$\displaystyle{S}_{{7}}={64}\frac{{{1}+\frac{{1}}{{128}}}}{{{1}+\frac{{1}}{{2}}}}$

$\displaystyle{S}_{{7}}={64}\cdot\frac{{43}}{{64}}$

$\displaystyle{S}_{{7}}={43}$

Therefore, $\displaystyle{\sum_{{{i}={1}}}^{{7}}}{64}{{\left(-\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}={43}$ ...

$\displaystyle{\sum_{{{i}={1}}}^{{7}}}{64}{{\left(-\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}$

The given summation notation has a form

$\displaystyle{\sum_{{{i}={1}}}^{{i}}}{a}_{{1}}{{\left({r}\right)}}^{{{i}-{1}}}$

Base on this, the first term and common ratio of the geometric series can be identified. The values are a1=64 and r=-1/2.

Plugging in the values of the a1 and r to the formula of geometric series

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

the sum of the first seven terms will be:

$\displaystyle{S}_{{7}}={64}\cdot\frac{{{1}-{{\left(-\frac{{1}}{{2}}\right)}}^{{7}}}}{{{1}-{\left(-\frac{{1}}{{2}}\right)}}}$

$\displaystyle{S}_{{7}}={64}\cdot{\left(\frac{{{1}-{\left(-\frac{{1}}{{128}}\right)}}}{{{1}-{\left(-\frac{{1}}{{2}}\right)}}}\right.}$

$\displaystyle{S}_{{7}}={64}\frac{{{1}+\frac{{1}}{{128}}}}{{{1}+\frac{{1}}{{2}}}}$

$\displaystyle{S}_{{7}}={64}\cdot\frac{{43}}{{64}}$

$\displaystyle{S}_{{7}}={43}$

Therefore, $\displaystyle{\sum_{{{i}={1}}}^{{7}}}{64}{{\left(-\frac{{1}}{{2}}\right)}}^{{{i}-{1}}}={43}$ .

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