A) 55
B) 66
C) 77
D) 88
E) 99
Correct Answer: B
Solution :
Given series, \[{{1}^{2}}-{{2}^{2}}+{{3}^{2}}-{{4}^{2}}+{{....11}^{2}}\] \[=({{1}^{2}}+{{3}^{2}}+{{5}^{2}}+....+{{11}^{2}})\] \[-({{2}^{2}}+{{4}^{2}}+...+{{10}^{2}})\] \[=({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{....11}^{2}})-2({{2}^{2}}+{{4}^{2}}\]\[+.....+{{10}^{2}})\] \[=({{1}^{2}}+{{2}^{2}}+...+{{11}^{2}})-{{2.2}^{2}}({{1}^{2}}+{{2}^{2}}\]\[+.....+{{5}^{2}})\] \[=11.\frac{(11+1)(22+1)}{6}\] \[-8.\frac{5(5+1)(10+1)}{6}\] \[=\frac{11.12.23}{6}-\frac{8.5.6.11}{6}\] \[\left[ \because \Sigma {{n}^{2}}=\frac{n(n+1)(2n+1)}{6} \right]\] \[=506-440\] \[=66\]
You need to login to perform this action.
You will be redirected in
3 sec
