The limit of the sum of an infinite number of terms in a geometric progression is (a)/(1-r), where a denotes the first term and -1< r<1 denotes the common ratio. The limit of the sum of their squares is:

1951 AMC 12 Problem 11

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11.

The limit of the sum of an infinite number of terms in a geometric progression is

\dfrac{a}{1-r},

where

a

denotes the first term and

-1\lt r\lt1

denotes the common ratio. The limit of the sum of their squares is:

\dfrac{a^2}{(1-r)^2}

\dfrac{a^2}{1+r^2}

\dfrac{a^2}{1-r^2}

\dfrac{4a^2}{1+r^2}

None of these

Squaring every term also squares the common ratio

The squared series begins

a^2+a^2r^2+a^2r^4+\cdots

The squared terms form a geometric series with first term

a^2

and ratio

r^2.

Since

|r|\lt1,

its sum is

\frac{a^2}{1-r^2}.

Thus, the correct answer is C.