What is the sum of the geometric series \(\sum_{n=1}^{10}6(2)^{n}\)

Solution:

The sum of the geometric series can be symbolically written as:

\(\sum_{n=1}^{10}6(2)^{n} = 6(2)^{1} + 6(2)^{2} + 6(2)^{3}+ + 6(2)^{10}\)

The above expression can also be written as:

\(\sum_{n=1}^{10}6(2)^{n} = 6[(2)^{1} + (2)^{2} + (2)^{3}+ + (2)^{10}]\)

The series with the brackets on the RHS of the above equation is written below as:

\((2)^{1} + (2)^{2} + (2)^{3}+ + (2)^{10}\)

The above is a geometric progression with a = 2 and r = 2 and n = 10. The sum of a geometric series is given as:

\(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\)

\(S_{10} = \frac{2(2^{10} - 1)}{2 - 1}\)

S_{10} = \(S_{10} = 2(2^{10} - 1)\)

= 2(1024 - 1)

= 2046

Therefore \(\sum_{n=1}^{10}6(2)^{n} = 6(2046)

= 12276

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What is the sum of the geometric series \(\sum_{n=1}^{10}6(2)^{n}\)

Summary:

The sum of the geometric series \(\sum_{n=1}^{10}6(2)^{n}\) is 12276