Question

If the sum of the first $15$ terms of the series
${\left(\frac{3}{4}\right)}^3+{\left(1\frac{1}{2}\right)}^3+{\left(2\frac{1}{4}\right)}^3+3^3+{\left(3\frac{3}{4}\right)}^3+....$

is equal to $225k$, then $k$ is equal to :

• A. $9$
• B. $27$
• C. $54$
• D. $108$

Answer 1

The correct option is B $27$

${\left(\frac{3}{4}\right)}^3+{\left(1\frac{1}{2}\right)}^3+{\left(2\frac{1}{4}\right)}^3+3^3+{\left(3\frac{3}{4}\right)}^3+....$

$={\left(\frac{3}{4}\right)}^3+{\left(\frac{3}{2}\right)}^3+{\left(\frac{9}{4}\right)}^3+3^3+{\left(\frac{15}{4}\right)}^3+....$

$={\left(\frac{3}{4}\right)}^3+{\left(\frac{6}{4}\right)}^3+{\left(\frac{9}{4}\right)}^3+{\left(\frac{12}{4}\right)}^3+{\left(\frac{15}{4}\right)}^3+....$

$={\left(\frac{3}{4}\right)}^3[1^3+2^3+3^3+...\text{upto}15\text{terms}]$
$={\left(\frac{3}{4}\right)}^3\times {\left(\frac{15\times 16}{2}\right)}^2$
$=\frac{27}{64}\times 225\times 64$
$=27\times 225$
$\therefore k=27$