| What is the value of \[[{{1}^{2}}-{{2}^{2}}+{{3}^{3}}-{{4}^{2}}+{{5}^{2}}-{{6}^{2}}+...+{{11}^{2}}-{{12}^{2}}]?\] |
A) 55
B) \[-78\]
C) \[-\,55\]
D) 78
Correct Answer: B
Solution :
| \[[{{1}^{2}}-{{2}^{2}}+{{3}^{2}}-{{4}^{2}}+{{5}^{2}}-{{6}^{2}}+....+{{11}^{2}}-{{12}^{2}}]\] |
| \[=\,\,[(1+2)(1-2)+(3+4)(3-4)+(5+6)\] |
| \[(5-6)+...+(11+12)(11-12)]\] |
| \[=[3\times -1+7\times -1+11\times -1+...+23\times -1]\] |
| \[=[-\,3-7-11-....-23]\] |
| This series is in AP with first term \[=-\,3\] |
| and last term \[=-\,23\] |
| \[\because \] \[l=a+(n-1)\,d\] |
| \[-\,23=-\,3+(n-1)\times -\,4\] |
| \[\Rightarrow \] \[-\,23=-\,3-4n+4\] |
| \[\Rightarrow \] \[4n=24\]\[\Rightarrow \]\[n=6\] |
| \[\therefore \] Sum \[=\frac{6}{2}(-\,3-23)=6\times -13=-78\] |
| Alternate Method |
| We can do it directly as, |
| \[[-\,3-7-11-15-19-23]=-78\] |
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