# Evaluating Sum at bounds

I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$

And have found the general equation

$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$

However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$

does not yield the correct answer, yet evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}$$

Im at a loss here, why am I not getting the correct answer by finding the difference of the sum between the two bounds?

• When I set $n=1$ and evaluate the two sides of your "general equation", I get $\sum_{r=2}^2 r(r+1) = \frac{2+6+4}{6}$ which simplifies to $6=2$. Jul 12, 2020 at 13:00

Let the terms of the sum be $$a_n$$. You need to find: $$a_{n+1}+a_{n+2}+\cdots+a_{2n}=\\ (a_1+\cdots+a_{n}+a_{n+1}+\cdots+a_{2n})-(a_1+\cdots+a_n)=\\ S_{2n}-S_n$$ In your first method, you are subtracting the term $$a_{n+1}$$ and losing it.

Addendum: Note the correct formula to use is: $$S_n=\sum_{k=1}^n k(k+1)=\frac{2n^3+6n^2+4n}{6}$$ Now consider the difference: $$\sum_{r=n+1}^{2n} r(r+1)=S_{2n}-S_n=\\ \frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}=\\ \frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$

• I do not understand the first statement, why is the sum = S2n-Sn , whereas the lower bound is n+1? Jul 12, 2020 at 14:49
• That is the reason. You do not want to lose it. You should keep it. So, you subtract all unneeded terms. Another example, say you need to add numbers from 5 to 10. First you find the sum from 1 to 10. Second you find the sum from 1 to 4 (not 5). Then you subtract the second sum from the first. Jul 12, 2020 at 15:07

Your general equation should be \begin {align} \sum_{r=n+1}^{2n} r(r+1)&=\sum_{r=n+1}^{2n} (r^2+r)\\ &=\sum_{r=1}^{2n} (r^2+r)-\sum_{r=1}^{n} (r^2+r)\\ &=\frac 16\left((2n)(2n+1)(4n+1)\right)+\frac 12\left(2n(2n+1)\right)-\frac 16\left((n)(n+1)(2n+1)\right)+\frac 12\left(n(n+1)\right)\\ &=\frac 16\left(16n^3+12n^2+2n\right)+\frac 12\left(4n^2+2n)\right)-\frac 16\left(2n^3+3n^2+n\right)+\frac 12\left(n^2+n\right)\\ &=\frac 16\left(14n^3+9n^2+n\right)+\frac 12\left(3n^2+n)\right)\\ &=\frac 16\left(14n^3+18n^2+4n\right) \end {align} which does not match your result and checks with Alpha.

• Can you please explain why you have evaluated it at 2n −n and not as 2n−(n+1) Jul 12, 2020 at 14:55
• Because $n+1$ is supposed to be included in the sum. The second sum is subtracting all the terms that are in the first sum but not on the left hand side. Jul 12, 2020 at 15:16

Another approach: it's clear that the result is a polynomial of $$n$$ of degree $$3$$, let $$\sum\limits_{r=n+1}^{2n}(r^2+r)=An^3+Bn^2+Cn+D=P(n)$$ thus \begin{align*}P(n)-P(n-1)&=\sum\limits_{r=n+1}^{2n}(r^2+r)-\sum\limits_{r=n}^{2n-2}(r^2+r)\\ &=-(n^2+n)+((2n-1)^2+(2n-1))+((2n)^2+2n)\\ &=7n^2-n\\ &\equiv A(3n^2-3n+1)+B(2n-1)+C\\ &=3An^2+(-3A+2B)n+(A-B+C) \end{align*} $$\begin{cases} 3A=7\\ -3A+2B=-1\\ A-B+C=0\\ A+B+C+D=P(1)=2^2+2=6 \end{cases}$$ $$\begin{cases} A=\frac{7}{3}\\ B=3\\ C=\frac{2}{3}\\ D=0 \end{cases}$$ Thus $$\sum\limits_{r=n+1}^{2n}(r^2+r)=\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$