JEE Main 12 Jan 2019 Evening Question 3
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 [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 27
B) 9
C) 108
D) 54
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ s={{( \frac{3}{4} )}^{3}}+{{( 1\frac{1}{2} )}^{3}}+{{( 2\frac{1}{4} )}^{3}}+3^{3}+{{( 3\frac{3}{4} )}^{3}} $ $ +…15 $ terms
$ ={{( \frac{3}{4} )}^{3}}+{{( \frac{3}{2} )}^{3}}+{{( \frac{9}{4} )}^{3}}+3^{3}+{{( \frac{15}{4} )}^{3}} $ $ +…15 $ terms
$ ={{( \frac{3}{4} )}^{3}}+{{( \frac{6}{4} )}^{3}}+{{( \frac{9}{4} )}^{3}}+{{( \frac{12}{4} )}^{3}}+{{( \frac{15}{4} )}^{3}} $ $ +…15 $ terms
$ =\frac{27}{64}\sum\limits _{x=1}^{15}{x^{3}}=\frac{27}{64}{{[ \frac{15(15+1)}{2} ]}^{2}}=27\times 225 $
$ \therefore $ $ 225k=27\times 225\Rightarrow k=27 $
