Notes: `sum_(n = 1)^7 4^(n-1)` Find the sum of the finite geometric series.

Sunday, June 12, 2011

The given geometric series is:

$\displaystyle{\sum_{{{n}={1}}}^{{7}}}{{4}}^{{{n}-{1}}}$

Take note that if the geometric series has a form:

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

its finite sum is:

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

Rewriting the given sigma notation in exact form as above, it becomes:

$\displaystyle{\sum_{{{n}={1}}}^{{7}}}{{4}}^{{{n}-{1}}}=\sum_{{{n}={1}}}{1}\cdot{{4}}^{{{n}-{1}}}$

From here, it can be seen that the values of the first term and the common ratio are a1=1 and r=4.

Plugging in the values of a1 and r to the formula of Sn, the sum of the first seven terms of the geometric sequence is:

$\displaystyle{S}_{{7}}={1}\cdot\frac{{{1}-{{4}}^{{7}}}}{{{1}-{4}}}={5461}$

Therefore, $\displaystyle{\sum_{{{n}={1}}}^{{7}}}{{4}}^{{{n}-{1}}}={5461}$ .

Notes