Question

If the sum of the first 15 terms of the series $⟮\frac{3}{4}{⟯}^3+⟮1\frac{1}{2}{⟯}^3+⟮2\frac{1}{4}{⟯}^3+3^3+⟮3\frac{3}{4}{⟯}^3+...........$ is equal to 225 k. then k is equal to:

• A. 9
• B. 27
• C. 108
• D. 54

Answer 1

The correct option is B 27

$⟮\frac{3}{4}{⟯}^3+⟮\frac{6}{4}{⟯}^3+⟮\frac{9}{4}{⟯}^3+⟮\frac{12}{4}{⟯}^3+..............15$ term

$S=\sum _{r=1}^{15}{\left[\frac{3r}{4}\right]}^3=\frac{27}{64}\sum r^3$

$=\frac{27}{64}\sum _{r=1}^{15}{r}^3$

$=\frac{27}{64}.[\frac{15(15+1)}{2}{]}^2$

$=27\times 225$ (given in question)

$K=27$