Sum of `n` terms the series : `1^2-2^2+3^2-4^2+5^2-6^2+....`

Options

A. `(n(n+1))/(2)`
B. `(-n(n+1))/(2)`
C. `(n(n-1))/(2)`
D. `(-n(n-1))/(2)`

Answer

A

Explanation

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)`