Had this question from some exam which goes like so:
Let $V_r$ denote the sum of first $r$ terms of an arithmetic progression whose first term is $r$ and the common difference is $(2r-1)$. Let $T_r = V_{r+1} – V_{r}$ and $Q_r = T_{r+1}-T_r$ for $r=1, 2, 3 \cdots$
The sum $V_1 +V_2 + \cdots + V_n$ is?
Answer: $\frac{1}{12}n(n+1)(3n^2+n+2)$ (just writing so you could know the terms the answer is in)
My line of reasoning was this:
For any given AP, I generalized a formula for the calculation of sum of sums of terms, $\sum_{i=1}^{n}S_i =\frac{n(n+1)}{6} (3a+nd-d)$
Where $a, n$ and $ d$ have their usual meanings.
So I figured it is just the same thing, plug in values, and get the answer but after a while, I realized that
Let $V_r$ denote the sum of first "$r$" terms of an arithmetic progression whose first term is "$r$"
This r has a unique condition to it so I could not use what I just derived. Could someone help me with this.
Best Answer
The approach for this question is straightforward First express $V_r$ as sum of an AP ,then sum up For $$V(r) a=r d= 2r -1 n= r$$ $$V(r) = (r/2)*[2r + (r-1)(2r-1)]$$ which simplifies to $$V(r) = (1/2)(2r^3-r^2+r)$$ now use standard summation formulas to get $$∑V(r) = (1/2){(2)[n^2*(n+1)^2/4] -(1/6)(n)(n+1)(2n+1) +(1/2)(n)(n+1)}$$ which simplifies to $$∑V(r) = (1/12)(n)(n+1)(3n^2 +n+2)$$