A) 7830
B) 7520
C) 7820
D) 7510
Correct Answer: C
Solution :
\[1+6+\frac{9({{1}^{2}}+{{2}^{2}}+{{3}^{2}})}{7}\,+\,\frac{12({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}})}{9}\]\[+\frac{15({{1}^{2}}+{{2}^{2}}+.....+{{5}^{2}})}{11}+.....15\,\,terms\]\[=\,\,\,\frac{3({{1}^{2}})}{3}+\frac{6({{1}^{2}}+{{2}^{2}})}{5}+\frac{9({{1}^{2}}+{{2}^{2}}+{{3}^{2}})}{7}\]\[+\,\,\frac{12}{9}\,({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}})+.......\] \[{{T}_{r}}=\frac{3r}{2r+1}({{1}^{2}}+{{2}^{2}}+....+{{r}^{2}})\] \[Tr\,\,=\,\,\frac{3r}{2r+1}\,\frac{r(r+1)(2r+1)}{6}\,=\,\frac{1}{2}{{r}^{2}}(r+1)\] Sum of n terms \[=\,\,\sum\limits_{r=1}^{n}{{{T}_{r}}\,\,=\,\,\frac{1}{2}\,\,\sum\limits_{r=1}^{n}{({{r}^{3}}+{{r}^{2}})}}\] \[=\,\,\,\frac{1}{2}\,\left[ \frac{{{n}^{2}}{{(n+1)}^{2}}}{4}+\frac{n(n+1)(2n+1)}{6} \right]\] Sum upto 15 terms \[\Rightarrow \] then put \[n=15\] \[\frac{1}{2}\left( \frac{{{(15\times 16)}^{2}}}{4}\,+\,\frac{15\times 16\times 31}{6} \right)\,\,=\,\,7820\]
You need to login to perform this action.
You will be redirected in
3 sec
