Let S = 109 + 1085+1075^2++25^107+15^108.

Question

Let $S = 109 + \dfrac{108}{5}+\dfrac{107}{5^2}+\cdots+\dfrac{2}{5^{107}}+\dfrac{1}{5^{108}}$. Then the value of $16S - (25)^{-54}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 2175

Accepted Answer

Using the standard technique $S - \dfrac{S}{5}$:
$$\frac{4S}{5} = 109 - \left(\frac{1}{5}+\frac{1}{5^2}+\cdots+\frac{1}{5^{109}}\right) = 109 - \frac14\left(1-5^{-109}\right).$$

$$16S = 20\left[109-\frac14+\frac14\cdot5^{-109}\right] = 2175 + 5\cdot5^{-109} = 2175+5^{-108}.$$

Since $(25)^{-54} = 5^{-108}$, these cancel exactly:

$16S-(25)^{-54} = 2175.$

Solutions · 1