Notes: `sum_(n = 1)^8 5(-5/2)^(n - 1)` Find the sum of the finite geometric series.

Tuesday, January 1, 2013

$\displaystyle{\sum_{{{n}={1}}}^{{8}}}{5}{{\left(-\frac{{5}}{{2}}\right)}}^{{{n}-{1}}}$ Find the sum of the finite geometric series.

The geometric series is:

$\displaystyle{\sum_{{{n}={1}}}^{{8}}}{5}{{\left(-\frac{{5}}{{2}}\right)}}^{{{n}-{1}}}$

The given summation notation is in the form

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

It can be seen that the first term and common ratio of the geometric sequence are a1=5 and r=-5/2.Plugging in these two values to the formula of finite geometric series

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

then, the sum of the first eight terms is:

$\displaystyle{S}_{{8}}={5}\cdot\frac{{{1}-{{\left(-\frac{{5}}{{2}}\right)}}^{{8}}}}{{{1}-{\left(-\frac{{5}}{{2}}\right)}}}=-\frac{{278835}}{{128}}$

Therefore, $\displaystyle{\sum_{{{n}={1}}}^{{8}}}{5}{{\left(-\frac{{5}}{{2}}\right)}}^{{{n}-{1}}}=-\frac{{278835}}{{128}}$ .

Notes

Tuesday, January 1, 2013

$\displaystyle{\sum_{{{n}={1}}}^{{8}}}{5}{{\left(-\frac{{5}}{{2}}\right)}}^{{{n}-{1}}}$ Find the sum of the finite geometric series.

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