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What is the sum of an infinite geometric series if the first term is 81 and the common ratio is 2⁄3?

1 Answer

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The terms in the series would be


81+81\left(\frac23\right)+81\left(\frac23\right)^2+81\left(\frac23\right)^3+\cdots

Consider the first
n terms of this series, and call it


S_n=81+81\left(\frac23\right)+81\left(\frac23\right)^2+\cdots+81\left(\frac23\right)^(n-1)

We have


\frac23S_n=81\left(\frac23\right)+81\left(\frac23\right)^2+81\left(\frac23\right)^3+\cdots+81\left(\frac23\right)^n

and subtracting this from
S_n gives


S_n-\frac23S_n=81-81\left(\frac23\right)^n


\frac13S_n=81\left(1-\left(\frac23\right)^n\right)


S_n=243\left(1-\left(\frac23\right)^n\right)

As
n\to\infty, the
\left(\frac23\right)^n term will converge to 0 and leave you with


\displaystyle\lim_(n\to\infty)S_n=243

User HanJeaHwan
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