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jmc

algebra junior

Problem

What is the sum of

(\frac{1}{3}) + (\frac{1}{3})^{2} + (\frac{1}{3})^{3} + (\frac{1}{3})^{4}

?

Solution

This 4-term geometric series has first term

a_{0} = \frac{1}{3}

and ratio

r = \frac{1}{3}

, so it has value

\frac{\frac{1}{3} ( 1 - ( \frac{1}{3} ) ^{4} )}{1 - \frac{1}{3}} = \frac{\frac{1}{3} ( 1 - ( \frac{1}{3} ) ^{4} )}{\frac{2}{3}} = \frac{1}{2} (1 - (\frac{1}{3})^{4}) = \frac{1}{2} (\frac{80}{81}) = \frac{40}{81} .

Final answer

\dfrac{40}{81}