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
asked 2021-01-06
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:
\(\displaystyle{S}_{{n}}={\frac{{{a}_{{1}}-{a}_{{1}}{r}^{{n}}}}{{{1}-{r}}}}\)

Answers (1)

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.
0

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