`Let S_n=Sigma_(k=1)^(4n) (-1)^((k(k+1))/2)k^2`.Then `S_n` can take value (s)

Options

A. 1056
B. 1088
C. 1120
D. 1332

Answer

A, D

Explanation

`S_(n)=sum_(k=1)^(4n)(-1)^(k(k+1))/2)k^(2)`
`=sum_(r-0)^((n-1))((4r+4)^(2)+(4r+3)^(2)-(4r+2)^(2)-(4r+1)^(2))`
`=sum_(r=0)^((n-1))(2(8r+6)+2(8r+4))`
`=sum_(r=0)^((n-1))(32r+20)`
`=16(n-1)n+20n`
`=4n(4n+1)`
`={{:(1056 " for n ="8),("1332 for n= 9"):}`