Expert Answer to "Find the sum of the convergent series. ∑n=0∞5(23)n"

Khadija Wells

Answered question

2021-02-15

Find the sum of the convergent series.

\sum_{n = 0}^{\infty} 5 (\frac{2}{3})^{n}

toroztatG

Skilled2021-02-16Added 98 answers

To find the sum of the convergent series:

\sum_{n = 0}^{\infty} 5 (\frac{2}{3})^{n}

Solution:
Expanding the given series, we get

\sum_{n = 0}^{\infty} 5 (\frac{2}{3})^{n} = 5 (\frac{2}{3})^{1} + 5 (\frac{2}{3})^{2} + 5 (\frac{2}{3})^{3} + . . .

= 5 [(\frac{2}{3})^{1} + (\frac{2}{3})^{2} + (\frac{2}{3})^{3} + . . .]

Now, taking the series

(\frac{2}{3})^{1} + (\frac{2}{3})^{2} + (\frac{2}{3})^{3} + . . .

Here,we can find that sequence is in geometric progression.
First term is

a_{1} = \frac{2}{3}

Common ratio is:

r = \frac{(\frac{2}{3})^{2}}{(\frac{2}{3})}

= \frac{2}{3}

Sum of infinite terms of G.P. is given as:

S = \frac{a_{1}}{1 - r}

Sum of the sequence

(\frac{2}{3})^{1} + (\frac{2}{3})^{2} + (\frac{2}{3})^{3} + . . .

will be:

S = \frac{\frac{2}{3}}{1 - \frac{2}{3}}

= \frac{\frac{2}{3}}{\frac{1}{3}}

= 2

Now, sum of the series

\sum_{n = 0}^{\infty} 5 (\frac{2}{3})^{n}

will be:

\sum_{n = 0}^{\infty} 5 (\frac{2}{3})^{n} = 5 [(\frac{2}{3})^{1} + (\frac{2}{3})^{2} + (\frac{2}{3})^{3} + . . .]

= 5 \cdot 2

= 10

Hence, required sum is 10.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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