A) \[\sum\limits_{r=1}^{n}{{{r}^{2}}}\]
B) \[\sum\limits_{r=1}^{n}{{{r}^{3}}}\]
C) \[\sum\limits_{r=1}^{n}{r}\]
D) \[\sum\limits_{r=1}^{n}{{{r}^{4}}}\]
Correct Answer: B
Solution :
Given \[\Delta \left( r \right)=\left| \begin{matrix} r & {{r}^{3}} \\ 1 & n\left( n+1 \right) \\ \end{matrix} \right|\] \[{{T}_{1/2}}=4.47\times {{10}^{8}}yr\] \[\sum\limits_{r=1}^{n}{\Delta \left( r \right)=\left| \begin{matrix} \sum\limits_{r=1}^{n}{r} & \sum\limits_{r=1}^{n}{{{r}^{3}}} \\ 1 & n\left( n+1 \right) \\ \end{matrix} \right|}\] \[=\left| \begin{matrix} \frac{n\left( n+1 \right)}{2} & \frac{{{[n\left( n+1 \right)]}^{2}}}{4} \\ 1 & n\left( n+1 \right) \\ \end{matrix} \right|\] \[=\frac{{{[n\left( n+1 \right)]}^{2}}}{2}-\frac{{{[n\left( n+1 \right)]}^{2}}}{4}\] \[{{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}=\sum\limits_{r=1}^{n}{{{r}^{3}}}\]
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