Let `V_(r )` denotes the sum of the first r terms of an arithmetic progression whose first term is r and the common difference is `(2r-1)`. Let `T_(r )=V_(r+1)-V_(r)-2" and " Q_(r )=T_(r+1)-T_(r )" for " r=1,2,"...."`
The sum V1​+V2​+...+Vn​ is

Options

A. `(1)/(12)n(n+1)(3n^(2)-n+1)`
B. `(1)/(12)n(n+1)(3n^(2)+n+2)`
C. `(1)/(2)n(2n^(2)-n+1)`
D. `(1)/(3)(2n^(3)-2n+3)`

Answer

B

Explanation

`V_(r)=(r )/(2)[(2r+(r-1)(2r-1))]=(1)/(2)(2r^(3)-r^(2)+r)`
`:.sum_(r=1)^(n)V_(r)=(1)/(2)[2sum_(r=1)^(n)r^(3)-sum_(r=1)^(n)r^(3)+sum_(r=1)^(n)r]`
`=(1)/(2)[2((n(n+1))/(2))^(2)-(n(n+1)(2n+1))/(6)+(n(n+1))/(2)]`
`=(1)/(2)n(n+1)(3n^(2)+n+2)`