A) odd and divisible by 5
B) divisible by 10
C) odd and not divisible by 5
D) None of the above
Correct Answer: A
Solution :
\[({{11}^{3}}+{{12}^{3}}+{{13}^{3}}+...+{{20}^{3}})\] \[=({{1}^{3}}+{{2}^{3}}\text{+}{{3}^{3}}+...+{{20}^{3}})-({{1}^{3}}\text{+}{{2}^{3}}+{{3}^{3}}+{{...10}^{3}})\] \[={{\left( \frac{20(20+1)}{2} \right)}^{2}}-{{\left( \frac{10(10+1)}{2} \right)}^{2}}\] \[\left[ \because \Sigma {{n}^{3}}=\left( \frac{n{{(n+1)}^{2}}}{2} \right) \right]\] \[={{(10.21)}^{2}}-{{(5.11)}^{2}}\] \[=(210-55)(210+55)\] \[=(155)(265)=41075\] which is odd and divisible by 5.
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