A) \[\frac{1}{24}\]
B) \[\frac{1}{12}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{3}\]
Correct Answer: A
Solution :
(r+1) th term of the series \[{{t}_{r+1}}=(r+1).\sum\limits_{k=1}^{n=r}{k}\] \[=\frac{(r+1)(n-r)(n-r+1)]}{2}\] \[=\frac{(r+1){{(n-r)}^{2}}+(n-r)]}{2}\] \[=\frac{r{{(n+r)}^{2}}+(n-r)]}{2}+\frac{{{(n+r)}^{2}}+(n-r)}{2}\] \[=\frac{r[{{(n+r)}^{2}}+(n-r)]}{2}P(n),\] Where \[P(n)=\frac{{{n}^{2}}-(2n-1)r+n+{{r}^{2}}]}{2}\] \[=\frac{r}{2}[{{r}^{2}}-(2n+1)r+{{r}^{2}}+n]+P(n)\] \[=\frac{r}{2}[{{r}^{3}}-(2+1){{r}^{2}}+({{n}^{2}}+n)r]+P(n)\] \[\therefore \]Sum of the series \[\sum\limits_{r=0}^{n-1}{{{t}_{r+1}}}\] \[=\frac{1}{2}\left[ \sum\limits_{r=0}^{n-1}{{{r}^{3}}-}(2n+1)\sum\limits_{r=0}^{n-1}{{{r}^{2}}}+n(n+1)\sum\limits_{r=0}^{n-1}{r} \right]+q(n)\] where \[q(n)=\sum\limits_{r=0}^{n-1}{p(n)}\]is a polynomial in n of degree 3. \[=\frac{1}{2}\left[ {{\left\{ \frac{(n-1)n}{2} \right\}}^{2}}-(2n+1)\frac{(n-1)n(1n-1)}{6}+n(n+1)\frac{(n-1)n}{2} \right]+q(n)\]\[=\frac{1}{2}\left[ \frac{({{n}^{2}}-2n+1)}{4}-\frac{({{n}^{2}}-n)(4{{n}^{2}}-1)}{6}+\frac{{{n}^{2}}({{n}^{2}}-1)}{2} \right]+q(n)\] \[\therefore \]Required limit \[=\frac{1}{2}\left( \frac{1}{4}-\frac{4}{6}+\frac{1}{2} \right)\] \[=\frac{1}{2}\left( \frac{3}{4}-\frac{2}{3} \right)=\frac{1}{24}\]
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