Sequence And Series Question 143
Question: $ \frac{1^{3}+2^{3}+3^{3}+4^{3}+{{……..12}^{3}}}{1^{2}+2^{2}+3^{2}+4^{2}+{{………12}^{2}}}= $
[MP PET 1998]
Options:
A) $ \frac{234}{25} $
B) $ \frac{243}{35} $
C) $ \frac{263}{27} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{1^{3}+2^{3}+3^{3}+4^{3}+………+12^{3}}{1^{2}+2^{2}+3^{2}+4^{2}+……..+12^{2}} $ $ =\frac{( \sum\limits_{n=1}^{12}{n^{3}} )}{( \sum\limits_{n=1}^{12}{n^{2}} )}={{[ \frac{n(n+1)}{2} ]}^{2}}\times \frac{6}{n,(n+1),(2n+1)} $ $ =\frac{3}{2}.\frac{n,(n+1)}{(2n+1)}=\frac{3}{2}.\frac{12.13}{25}=\frac{234}{25} $ , [Putting $ n=12 $ ].
BETA
