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In question 62,they have solved the question by subtracting Vr+1 and Vr and then subtracting 2 to obtain 3r^2 +2r-1. I have fully understood this.But what I did not understand is that why the equation is not obtained if I equate the difference of Vr+1 and Vr to the r+1th term of the original AP.Please help.

Note:I am not interested in this problem in particular. I want to understand when and where the rule (that difference in sum of r+1 terms and r terms of an AP always equal to the r+1st term) does not hold good.So please don't close this question saying that i should show some effort to solve the problem.

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$$\left(\sum\limits_{k=1}^{n+1}a_k\right) - \left(\sum\limits_{k=1}^n a_k\right)=\left(a_{n+1}+\sum\limits_{k=1}^{n}a_k\right) - \left(\sum\limits_{k=1}^n a_k\right)=a_{n+1}$$

This is true regardless of what the sequence $a_n$ actually is, be it an arithmetic progression or otherwise.

In your particular problem however, $V_{r+1}$ and $V_r$ are not both sums of the same arithmetic progression. They are both sums of arithmetic progressions, yes, but with different starting values and different stepsizes. Your problem defines $V_r = \sum\limits_{k=0}^{r-1} (r+k(2r-1))$, (i.e. sum of $r$ terms in arithmetic progression with first term $r$ and stepsize $2r-1$) so in this case $V_{r+1}-V_r$ does not fit the form of the above identity.