Question:
If the sum of the first 15 tems of the $\operatorname{series}\left(\frac{3}{4}\right)^{3}+\left(1 \frac{1}{2}\right)^{3}+\left(2 \frac{1}{4}\right)^{3}+3^{3}+\left(3 \frac{3}{4}\right)^{3}+\ldots$ is
equal to $225 \mathrm{k}$, then $\mathrm{k}$ is equal to :
Correct Option: , 2
Solution:
$\mathrm{S}=\left(\frac{3}{4}\right)^{3}+\left(\frac{6}{4}\right)^{3}+\left(\frac{9}{4}\right)^{3}+\left(\frac{12}{4}\right)^{3}+\ldots \ldots \ldots \ldots 15$ term
$=\frac{27}{64} \sum_{r=1}^{15} r^{3}$
$=\frac{27}{64} \cdot\left[\frac{15(15+1)}{2}\right]^{2}$
$=225 \mathrm{~K}$ (Given in question)
$K=27$
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