If the sum of the first 15 terms of the series 34^3+112^3+214^3+3^3+334^3+ is equal to 225…

Question

If the sum of the first 15 terms of the series $\dfrac{3}{4^3}+1\dfrac{1}{2^3}+2\dfrac{1}{4^3}+3^3+3\dfrac{3}{4^3}+\cdots$ is equal to $225k$, then $k$ is equal to:

Accepted Answer

The series is $\dfrac{3}{4^3}+\dfrac{6}{4^3}+\dfrac{9}{4^3}+\dfrac{12}{4^3}+\cdots$, general term $T_r = \dfrac{3r}{4^3}\cdot r^2$... i.e. effectively $T_r=\dfrac{3r^3}{4^3}$ after regrouping.

$$225k = \sum_{r=1}^{15}T_r = \frac{3}{64}\left(\frac{15\times16}{2}\right)^2 = \frac{27}{64}\times14400 \cdots \implies k=27.$$

Solutions · 1