Question

Use the formula to evaluate the series 6+2+2/3+2/9

Answer 1

If you're talking about adding just these four terms, notice that


`S_4=6+2+\frac23+\frac29`

`S_4=6\left(1+\frac13+\frac1{3^2}+\frac1{3^3}\right)`

`\frac13S_4=6\left(\frac13+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}\right)`


`image`

`S_4=9*(80)/(81)`

`S_4=\frac{80}9`

If you were referring to an infinite sum, so that the pattern continues, note that the same procedure as above can be applied to the
`n`th partial sum:


`S_n=6\left(1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}\right)`

`\frac23S_n=6\left(1-\frac1{3^(n+1)}\right)`

`S_n=9\left(1-\frac1{3^(n+1)}\right)`

Then as
`n\to\infty`, the exponential term approaches 0, leaving you with


`6+2+\frac23+\frac29+\cdots=9`