The sum [63^26 + 1013^25 + 1023^24 + 102^23^23 + + 102^243^3] is equal to:

Question

The sum
$$\frac{6}{3^{26}} + \frac{10\cdot1}{3^{25}} + \frac{10\cdot2}{3^{24}} + \frac{10\cdot2^2}{3^{23}} + \cdots + \frac{10\cdot2^{24}}{3^{3}}$$
is equal to:

Accepted Answer

Let $S' = \dfrac{10}{3^{25}} + \dfrac{10\cdot2}{3^{24}} + \cdots + \dfrac{10\cdot2^{24}}{3^{3}} = \dfrac{10}{3^{25}}\sum_{k=0}^{24}(2\cdot3)^k = \dfrac{10}{3^{25}}\sum_{k=0}^{24}6^k.$

$$\sum_{k=0}^{24}6^k = \frac{6^{25}-1}{5} \implies S' = \frac{10}{3^{25}}\cdot\frac{6^{25}-1}{5} = \frac{2}{3^{25}}\left(6^{25}-1\right) = 2\cdot2^{25} - \frac{2}{3^{25}} = 2^{26}-\frac{2}{3^{25}}.$$

Since $\dfrac{2}{3^{25}} = \dfrac{6}{3^{26}}$, adding the first term of the original series:

$$S = \frac{6}{3^{26}} + S' = \frac{6}{3^{26}} + 2^{26} - \frac{6}{3^{26}} = 2^{26}.$$

Solutions · 1