# If the sum of the first 15 terms of the series $\left ( \frac{3}{4} \right )^{3}+\left ( 1\frac{1}{2} \right )^{3}+\left ( 2\frac{1}{4} \right )^{3}+3^{3}+\left ( 3\frac{3}{4} \right )^{3}+\cdots$is equal to 225 k, then k is equal to : Option 1)  $9$Option 2)  $27$Option 3)  $108$Option 4)  $54$

Summation of series of natural numbers -

$\sum_{k=1}^{n}K= \frac{1}{2}n\left ( n+1 \right )$

- wherein

Sum of first n natural numbers

$1+2+3+4+------+n= \frac{n(n+1)}{2}$

Summation of series of natural numbers -

$\sum_{k=1}^{n}K^{3}=\left ( \sum_{k=1}^{n}K \right )^{2}$

- wherein

$1^{3}+2^{3}+3^{3}+-------n^{3}\\= \left ( 1+2+3+-----+n \right )^{2}$

$S=\sum_{r=1}^{15}\left ( \frac{3r}{4} \right )^{3}=\frac{27}{64}\sum r^{3}$

$=\frac{27}{64}\times \frac{\left ( 15\times 16 \right )^{2}}{4}$

$=27\times 225$

$K=27$

Option 1)

$9$

Option 2)

$27$

Option 3)

$108$

Option 4)

$54$

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