A) \[A\xrightarrow[{}]{{}}B;{{k}_{1}}={{10}^{10}}{{e}^{-20,000/T}}\]
B) \[C\xrightarrow[{}]{{}}D;{{k}_{2}}={{10}^{12}}{{e}^{-24,606/T}}\]
C) \[{{k}_{1}}\]
D) \[{{k}_{2}}\]
Correct Answer: A
Solution :
Let \[{{T}_{r}}\]be the rth term of the given series. Then,\[{{T}_{r}}=\frac{2r+1}{{{1}^{2}}+{{2}^{2}}+...+{{r}^{2}}}=\frac{2r+1}{\frac{r(r+1)(2r+1)}{6}}\] \[=\frac{6}{r(r+1)}\]\[\Rightarrow \]\[{{T}_{r}}=6\left( \frac{1}{r}-\frac{1}{r+1} \right)\] So, the required sum is given by \[\sum\limits_{r=1}^{n}{{{T}_{r}}=6}\sum\limits_{r=1}^{n}{\left( \frac{1}{r}-\frac{1}{r+1} \right)}\] \[=6\left[ \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1} \right]\] \[=6\left[ 1-\frac{1}{n+1} \right]=\frac{6n}{n+1}\]
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