The sum of series `Sigma_(r=0)^(r) (-1)^r(n+2r)^2` (where n is even) is

Options

A. `-n^2+2n`
B. `-4n^2+2n`
C. `-n^2+3n`
D. `-n^2+4n`

Answer

B

Explanation

`sum_(r=0)^(n)(-1)^(r )(n+2r)^(2)`
`=n^(2)+(n+2)^(2)+(n+4)^(2)-…-(3n)^(2)`
`=(n-n-2)(n+n+2)+(n+4-n-6)(n+4+n+6)+…`
`=(-2)[(2n+2)+(2n+10)+(2n+18)+..+n/2` terms `]`
`=(-2)[2n(n//2)+(2+10+18+..+n//2` terms `]`
`=(-2)[n^(2)+n/4(4+(n/2-1)8)]`
`=(-2)[n^(2)+n+n^(2)-2n]`
`=-4n^(2)+2n`