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Using the fact that $\sum_{k=0}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}$, give a closed form sum for:

$\sum_{k=2n+1}^{3n}k(k+1)$.

So I made the substitution $i=k-(2n+1)$, and the new summation I wrote is:

$\sum_{i=0}^{n-1}(i+2n+1)(i+2n+2)$.

How do I give a closed form sum for this?

My guess is $2n+1$ is just a constant so if we define $A:=2n+1$, we have $\sum_{i=0}^{n-1}(i+A)(i+A+1)$, which kind of looks of the form $\sum_{k=0}^{n}k(k+1)=\frac{n(n+1)(n+2)}{3}$. Where do I go from here? I tried expanding the polynomial and separated the summation across the monomials but I end up with summations like $\sum_{i=0}^{n-1}A^2$ which I don't know how to evaluate since the index of summation is not present.

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    $\begingroup$ You are over-complicating things. If $\sum_{k=1}^{n}g(k) = f(n)$, then $$\sum_{k=2n+1}^{3n}g(k) = f(3n)-f(2n).$$ $\endgroup$ – Jack D'Aurizio Oct 26 '17 at 3:33
  • $\begingroup$ How did you figure out this generalization? I'm trying to work it out by hand to see how it works but I'm getting nowhere $\endgroup$ – VV6570 Oct 26 '17 at 3:50
  • $\begingroup$ It can be seen as the fundamental theorem of Calculus, $\int_{a}^{b}g(x)\,dx = G(b)-G(a)$. I guess one just needs a bit of confidence in manipulating sums and integrals, before such tricks become natural. $\endgroup$ – Jack D'Aurizio Oct 26 '17 at 3:58
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$$\sum_{k=2n+1}^{3n}k(k+1)=\sum_{k=0}^{3n}k(k+1)-\sum_{k=0}^{2n}k(k+1)$$

$$\sum_{k=2n+1}^{3n}k(k+1)=\frac{3n(3n+1)(3n+2)}{3}-\frac{2n(2n+1)(2n+2)}{3}$$

$$\sum_{k=2n+1}^{3n}k(k+1)=.......$$

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Using the form you derived, we have $$\sum_{i=0}^{n-1}(i+2n+1)(i+2n+2)=2\sum_{i=0}^{n-1}\binom {2n+2+i}2\\=2\left[ \binom {3n+2} 3-\binom {2n+2}3\right]\\ =\frac 13 n(19n^2+15n+2)$$

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