infinite geo series

[xc630]

Hello would someone please check my answer on the sum of this infinite geometric series.

SQRT(27)+ SQRT(9)+SQRT(3)+...

By puting the second term over the first term I foudn the ratio to be SQRT(3)/3.

So then I did the formula for the sum, 1st term/ 1-Ratio

SQRT27/ 1- (SQRT(3)/3)

I got 9SQRT(3)/ 3-SQRT(3).

I'm not sure if I did all the algebra right. Can someone check this for me?

 

[soroban]

Hello, xc630!

Your work looks fine to me . . .

\(\displaystyle \L\sqrt{27}\,+\,\sqrt{9}\,+\,\sqrt{3}\,+\,\cdots\)

By puting the second term over the first term, I found the ratio to be: \(\displaystyle \,\frac{\sqrt{3}}{3}\)

So then I did the formula for the sum: 1st term/(1 - Ratio)

\(\displaystyle \L\;\;\frac{\sqrt{27}}{1\,-\,\frac{\sqrt{3}}{3}}\)

I got: \(\displaystyle \L\,\frac{9\sqrt{3}}{3\,-\,\sqrt{3}}\;\;\) . . . Right!

I would rationalize it, though . . .

\(\displaystyle \L\;\;\frac{9\sqrt{3}}{3\,-\,\sqrt{3}}\,\cdot\,\frac{3\,+\,\sqrt{3}}{3\,+\,\sqrt{3}} \;= \;\frac{9\sqrt{3}(3\,+\,\sqrt{3})}{9\,-\,3}\;=\;\frac{27\sqrt{3}\.+\.27}{6}\;=\;\frac{27(1\.+\,\sqrt{3})}{6}\;=\;\frac{9(1\,+\,\sqrt{3})}{2}\)