Find the sum Σ k² × (1/3)^k over k = 1, 2, 3, …. Give the answer as a fraction in lowest terms.

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Squares with geometric weights

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Find the sum $\sum k^{2} \times (\frac{1}{3})^{k}$ over $k = 1, 2, 3, \dots$ . Give the answer as a fraction in lowest terms.

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Hint

Apply the operation of differentiating and then multiplying by $x$ to the geometric series, more than once.

Solution

Starting from $\sum x^{k} = \frac{1}{1 - x}$ and applying $x \frac{d}{d x}$ once gives $\sum k x^{k} = \frac{x}{( 1 - x ) ^{2}}$ .
Applying it again gives $\sum k^{2} x^{k} = \frac{x ( 1 + x )}{( 1 - x ) ^{3}}$ .
At $x = \frac{1}{3}$ , the numerator is $\frac{1}{3} \times \frac{4}{3} = \frac{4}{9}$ and the denominator is $\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$ .
So the sum is $\frac{\frac{4}{9}}{\frac{8}{27}} = \frac{3}{2}$ .

Answer: $\frac{3}{2}$