I have the following nested sum : $$\sum_{i=1}^{n}\sum_{j=1}^{i}\sum_{k=1}^{j}x = x+1$$

I don't have a clue how to solve this one, can somebody help me?

Thanks in advance. ${}{}$

  • 2
    $\begingroup$ I"m confused. Is it really supposed to be an $x$ on both sides here? $\endgroup$ – Zev Chonoles Feb 24 '13 at 13:09
  • 2
    $\begingroup$ This can't be correct. The right-hand side must depend on $n$, it seems to me. $\endgroup$ – bubba Feb 24 '13 at 13:13

I will use the rules found on this page.

First, note that $x$ is not in the base or limit of any of the sums, so it can be "pulled out": $$x\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j 1= x+1$$

Work from inside to outside, simplifying the sum: $$\sum_{k=1}^j 1 = j$$ $$\sum_{j=1}^ij = \frac{i(i+1)}{2}=\frac{1}{2}\cdot\left(i^2 + i\right)$$ $$\sum_{i=1}^n\frac{1}{2}\cdot\left(i^2 + i\right) = \frac{1}{2}\left(\sum_{i=1}^ni^2+\sum_{i=1}^ni\right) = \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right)$$

Now we're done, and can plug back in:

$$\frac{x}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right)=x+1$$

Now, solving for $x$: $$x\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)=1$$ $$ \begin{align*} x &= \frac{1}{\left(\frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} - 1\right)} \\ &= \frac{6}{n^3 + 3 n^2 + 2 n - 6} \end{align*} $$

  • $\begingroup$ Thank you very much, this made me understand it. $\endgroup$ – Energyfellow Feb 24 '13 at 20:42

After a little rearrangement we have

$$\begin{align*} 1+\frac1x&=\sum_{i=1}^n\sum_{j=1}^i\sum_{k=1}^j1\\ &=\sum_{i=1}^n\sum_{j=1}^ij\\ &=\sum_{1=1}^n\frac{i(i+1)}2\\ &=\frac12\left(\sum_{i=1}^ni^2+\sum_{i=1}^ni\right)\\ &=\frac12\left(\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2\right)\\ &=\frac1{12}\Big(n(n+1)(2n+1)+3n(n+1)\Big)\\ &=\frac{n(n+1)}{12}(2n+4)\\ &=\frac16n(n+1)(n+2)\\ &=\binom{n+2}3\;, \end{align*}$$


$$\begin{align*} x&=\frac1{\binom{n+2}3-1}\\ &=\frac6{n(n+1)(n+2)-6}\\ &=\frac6{n^3+3n^2+2n-6}\\ &=\frac6{(n-1)(n^2+4n+6)}\;. \end{align*}$$

  • $\begingroup$ Can't this be done combinatorially? $\endgroup$ – Ishan Banerjee Feb 24 '13 at 17:20
  • $\begingroup$ @Ishan: Conceivably, but I don’t at the moment see such an argument. $\endgroup$ – Brian M. Scott Feb 24 '13 at 17:50

This question can be solved in a different way. Your equation is equivalent to $1+\frac1x=\sum_{(i,j,k)\in A}1=|A|$

Where $A={(i,j,k)| 1\leq k \leq j\leq i\leq n }$

So,$|A|= ^nC_3 +2^nC_2+^nC_1$ This is because, we can choose 3 distinct i,j,k or distinct i with the same j and k, or distinct k with the same i and j, or i=j=k.

So we get $1+\frac1x=^nC_3 +2^nC_2+^nC_1=^{n+2}C_3$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.