Find the sum of n terms of the series `ab+(a-1)(b-1)+(a-2)(b-2)+....` if `ab=1/6` and `a+b=1/3`.

Options

A. `(n)/(6)(1-2n)^(2)`
B. `(n)/(6)(1+n-2n^(2))`
C. `(n)/(6)(1-2n+2n^(2))`
D. None of these

Answer

C

Explanation

`(c )` `S=ab+[ab+(a+b)+1]+[ab+2(a+b)+2^(2)]+…+[ab+(n-1)(a+b)+(n-1)^(2)]`
`=nab+(a+b)sum_(r=1)^(n-1)r+sum_(r=1)^(n-1)r^(2)`
`=nab+(a+b)(n(n-1))/(2)+((n-1)(n)(2n-1))/(6)`
`=(n)/(6)[1+(n-1){1+2n-1}]`
`=(n)/(6)[1+2n(n-1)]=(n)/(6)(1-2n+2n^(2))`