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,"...."`
`T_(r )` is always

Options

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

Answer

B

Explanation

We have,
`V_(r)=(r)/(2){2r+(r-1)(2r-1)}`
`rArr" "V_(r)=(r)/(2)[2r+2r^(2)-3r+1]=(r)/(2)(2r^(3)-r^(2)+r)`
`:." "V_(1)+V_(2)+V_(3)+. . . .+V_(n)`
`=underset(r=1)overset(n)sumV_(r)`
`(1)/(2)underset(r=1)overset(n)sum(2r^(3)-r^(2)+r)` ltbr `(1)/(2){2underset(r=1)overset(n)sumr^(3)-underset(r=1)overset(n)sumr^(2)+underset(r=1)overset(n)sumr}`
`=(1)/(2){:[2{(n(n+1))/(2)}^(2)-(n(n+1)(2n+1))/(6)+(n(n+1))/(2)]:}`
`=(1)/(12)n(n+1)(3n^(2)+n+2)`