If n is an odd integer greater than or equal to 1, then the value of `n^3 - (n-1)^3 + (n-2)^3 - (n-3)^3 + .... + (-1)^(n-1) 1^3`

Options

A. `((n+1)^(2)(2n-1))/(4)`
B. `((n-1)^(2)(2n-1))/(4)`
C. `((n+1)^(2)(2n+1))/(4)`
D. None of these

Answer

A

Explanation

Given that n is an odd integer greater than or equal to 1.
`S_(n)=n^(3)-(n-1)^(3)+(n-2)^(3)-"..."+(-1)^(n-1)1^(3)`
`=1^(3)-2^(3)+"...."+(n-2)^(3)-(n-1)^(3)+n^(3)`
[`:.` n is odd integer , so (n-1) is even integer]
`=(1^(3)-2^(3)+"...."+n^(3))-2*2^(3)(1^(3)-2^(3)+"...."+(n-1)/(2) " terms ")`
`=[(n(n+1))/(2)]^(2)-16*[((n-1)/(2)((n-1)/(2)+1))/(2)]^(2)`
`=(n^(2)(n+1)^(2))/(4)-(4(n-1)^(2)(n+1)^(2))/(16)=((n+1)^(2))/(4)[n^(3)-(n-1)^(2)]`
`=(n-1)^(2)/(4)*(2n-1)(1)=((2n-1)(n+1)^(2))/(4)`.