Let Vr denotes the sum of first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (2r-1).  Let Tr = Vr+1-Vr-2Tr is always:

The given arithmetic progression (A.P.) has its first term as 'r' and the common difference as '2r-1'.

The nth term of an A.P. is given by a + (n-1)d, where 'a' is the first term and 'd' is the common difference.

So, the rth term of the given A.P. would be r + (r-1)(2r-1) = r + 2r^2 - r - 2r + 1 = 2r^2 - r + 1.

Now, the sum of first 'n' terms of an A.P. is given by n/2[2a + (n-1)d].

So, the sum of first 'r' terms, Vr, of the given A.P. would be r/2[2r + (r-1)(2r-1)] = r/2[2r + 2r^2 - r - 2r + 1] = r/2[2r^2 - r + 1] = r^3 - r^2/2 + r/2.

Now, Tr = Vr+1 - Vr.

Substituting the expression for Vr in the above equation, we get Tr = [(r+1)^3 - (r+1)^2/2 + (r+1)/2] - [r^3 - r^2/2 + r/2] = r^3 + 3r^2 + 3r + 1 - r^2/2 - r/2 - 1/2 - r^3 + r^2/2 - r/2 = 3r^2 + 3r + 1/2.

So, Tr is always equal to 3r^2 + 3r + 1/2.