Find the sum to `n` terms of the series `1^2+2^2+3^2-4^2+5^2-6^2+. . . .`

Answer

When n in even, `-(n(n+1))/(2)`; When n is odd, `(n(n+1))/(2)`

Explanation

Clearly, nth term of the given series is negative or positive accordingly as n is even or odd, respectively.
(a) n is even:
`1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+…+(n-1)^(2)-n^(2)`
`=(1^(2)-2^(2))+(3^(2)-4^(2))+(5^(2)-6^(2))+…+((n-1)^(2)-n^(2))`
`=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+((n-1)-(n))(n-1+n)`
`=-(1+2+3+4+..+(n-1)+n)`
`=-(n(n+1))/2`
(b) n is odd:
`(1^(2)-2^(2))+(3^(2)-4^(2))+...+{(n-2)^(2)-(n-1)^(2)}+n^(2)`
`=(1-2)(1+2)+(3-4)(3+4)+...+[(n-2)-(n-1)][(n-2)+(n-2)]+n^(2)`
`=-(1+2+3+4+..+(n-2)+(n-1)+n^(2)`
`=-((n-1)(n-1+1))/2+n^(2)`
`=(n(n+1))/2`