A) \[\frac{234}{25}\]
B) \[\frac{243}{35}\]
C) \[\frac{263}{27}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+.........+{{12}^{3}}}{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........+{{12}^{2}}}\] \[=\frac{\left( \sum\limits_{n=1}^{12}{{{n}^{3}}} \right)}{\left( \sum\limits_{n=1}^{12}{{{n}^{2}}} \right)}={{\left[ \frac{n(n+1)}{2} \right]}^{2}}\times \frac{6}{n\,(n+1)\,(2n+1)}\] \[=\frac{3}{2}.\frac{n\,(n+1)}{(2n+1)}=\frac{3}{2}.\frac{12\,\,.\,\,13}{25}=\frac{234}{25}\], [Putting\[n=12\]].
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