Let S = 2+67+127^2+207^3+307^4+.

Question

Let $S = 2+\dfrac67+\dfrac{12}{7^2}+\dfrac{20}{7^3}+\dfrac{30}{7^4}+\cdots$. Then $4S$ is equal to:

Accepted Answer

The numerators $2,6,12,20,30,\ldots = n(n+1)$. Using the successive-subtraction technique twice (multiplying by $\tfrac17$ and subtracting) reduces $S$ to a geometric series:

$$\frac{36}{49}S = 2\left(\frac{1}{1-1/7}\right) = \frac{7}{3} \implies S = 2\left(\frac76\right)^3.$$

$$4S = 8\left(\frac76\right)^3 = \frac{8\times343}{216} = \frac{343}{27} = \frac{7^3}{3^3}.$$

Solutions · 1