A) 27
B) 9
C) 108
D) 54
Correct Answer: A
Solution :
Let\[s={{\left( \frac{3}{4} \right)}^{3}}+{{\left( 1\frac{1}{2} \right)}^{3}}+{{\left( 2\frac{1}{4} \right)}^{3}}+{{3}^{3}}+{{\left( 3\frac{3}{4} \right)}^{3}}\]\[+...15\]terms \[={{\left( \frac{3}{4} \right)}^{3}}+{{\left( \frac{3}{2} \right)}^{3}}+{{\left( \frac{9}{4} \right)}^{3}}+{{3}^{3}}+{{\left( \frac{15}{4} \right)}^{3}}\]\[+...15\]terms \[={{\left( \frac{3}{4} \right)}^{3}}+{{\left( \frac{6}{4} \right)}^{3}}+{{\left( \frac{9}{4} \right)}^{3}}+{{\left( \frac{12}{4} \right)}^{3}}+{{\left( \frac{15}{4} \right)}^{3}}\]\[+...15\]terms \[=\frac{27}{64}\sum\limits_{x=1}^{15}{{{x}^{3}}}=\frac{27}{64}{{\left[ \frac{15(15+1)}{2} \right]}^{2}}=27\times 225\] \[\therefore \]\[225k=27\times 225\Rightarrow k=27\]
You need to login to perform this action.
You will be redirected in
3 sec
