1950 AMC 12 Problem 43

Question

The sum to infinity of

\dfrac17+\dfrac{2}{7^2}+\dfrac{1}{7^3}+\dfrac{2}{7^4}+\cdots

is:

\dfrac15

\dfrac1{24}

\dfrac5{48}

\dfrac1{16}

None of these

Group the series into consecutive pairs of terms

Each pair is

\frac{1}{49}

times the preceding pair

Grouping consecutive terms gives a geometric series whose first grouped term is

\frac17+\frac2{49}=\frac9{49}

and whose ratio is

\tfrac1{49}.

Hence the sum is

\frac{\frac9{49}}{1-\frac1{49}} =\frac9{48} =\frac3{16}.

This is not among choices A through D.

Thus, the correct answer is E.

Problem 43 in other years