2
$\begingroup$

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 $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}$$

gives the correct answer

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

$\endgroup$
1
  • $\begingroup$ 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 $6=2$. $\endgroup$
    – Lee Mosher
    Jul 12, 2020 at 13:00

3 Answers 3

1
$\begingroup$

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: $$\sum_{r=n+1}^{2n} r(r+1)=S_{2n}-S_n=\\ \frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}=\\ \frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$

$\endgroup$
2
  • $\begingroup$ I do not understand the first statement, why is the sum = S2n-Sn , whereas the lower bound is n+1? $\endgroup$ Jul 12, 2020 at 14:49
  • 1
    $\begingroup$ 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. $\endgroup$
    – farruhota
    Jul 12, 2020 at 15:07
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Can you please explain why you have evaluated it at 2n −n and not as 2n−(n+1) $\endgroup$ Jul 12, 2020 at 14:55
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 12, 2020 at 15:16
0
$\begingroup$

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 $$\sum\limits_{r=n+1}^{2n}(r^2+r)=\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .