# Find the sum of the followinf series. Round to the nearest hundredth if necessary. 6+12+24+...+6144 Sum of a finite geometric series: S_n=frac{a_1-a_1r^n}{1-r}

Question
Series
Find the sum of the followinf series. Round to the nearest hundredth if necessary.
$$\displaystyle{6}+{12}+{24}+\ldots+{6144}$$
Sum of a finite geometric series:
$$\displaystyle{S}_{{n}}={\frac{{{a}_{{1}}-{a}_{{1}}{r}^{{n}}}}{{{1}-{r}}}}$$

2021-01-07
Given Data:
Series: $$\displaystyle{6}+{12}+{24}+\ldots+{6144}$$
The common ratio of the series is,
$$\displaystyle{r}={\frac{{{a}_{{2}}}}{{{a}_{{1}}}}}$$
Here, $$\displaystyle{a}_{{1}}$$ is the first term of series and $$\displaystyle{a}_{{2}}$$ is the second term of the series.
Substitute the values in the above equation.
$$\displaystyle{r}={\frac{{{12}}}{{{6}}}}$$
$$\displaystyle={2}$$
The last term of the series is,
$$\displaystyle{L}={a}_{{1}}{r}^{{{n}-{1}}}$$
Substitute the values in the above equation.
$$\displaystyle{6144}={6}{\left({2}\right)}^{{{n}-{1}}}$$
$$\displaystyle{2}^{{{n}-{1}}}={1024}$$
$$\displaystyle{2}^{{{n}-{1}}}={2}^{{{10}}}$$
$$\displaystyle{n}-{1}={10}$$
$$\displaystyle{n}={11}$$
The sum of finite geometric series is,
$$\displaystyle{S}_{{n}}={\frac{{{a}_{{1}}-{a}_{{1}}{r}^{{n}}}}{{{1}-{r}}}}$$
Here, n is the number of terms in the series.
Susbtitute the values in the above equation.
$$\displaystyle{S}_{{n}}={\frac{{{6}-{6}{\left({2}\right)}^{{{11}}}}}{{{1}-{2}}}}$$
$$\displaystyle{S}_{{n}}={\frac{{{6}-{6}{\left({2048}\right)}}}{{{1}-{2}}}}$$
$$\displaystyle{S}_{{n}}={\frac{{-{12282}}}{{-{1}}}}$$
$$\displaystyle={12282}$$
Thus, the sum of a finite geometric series is 12282.

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