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Sum of `n`terms the series : `1^2-2^2+3^2-4^2+5^2-6^2+`A. `(n(n+1))/(2)`B. `(-n(n+1))/(2)`C. `(n(n-1))/(2)`D. `(-n(n-1))/(2)` |
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Answer» Correct Answer - A If n is odd, then required sum is given by `1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+ . . . .+{(n-2)^(2)-(n-1)^(2)}+n^(2)` `=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+ . . . .+{(n-2)-(n-1)}{(n-2)+(n-1)}+n^(2)` `=-{1+2+3+ . . . +(n-2)+(n-1)}+n^(2)` `=-(n(n-1))/(2)+n^(2)=(n(n+1))/(2)` |
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