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I would like to know how it is to prove this problem:

$$\sum_{j = 1}^{n} j(j + 1) = \frac{1}{3}n(n + 1)(n + 2) \text{where } n \geq 1 \\ $$

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    $\begingroup$ Split it into the sums $\sum j^2$ and $\sum j$. $\endgroup$ – Dietrich Burde Apr 9 at 18:10
  • $\begingroup$ Have you considered using induction? $\endgroup$ – JacobCheverie Apr 9 at 18:54
  • $\begingroup$ @JacobCheverie I have to solve the problem by induction. $\endgroup$ – Chris Michael Apr 9 at 18:57
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Using induction, prove the case $n = 1:$

If $n = 1$, we wish to show $$\sum_{j = 1}^{1}j(j + 1) = \dfrac{1}{3}(1)(1+1)(1+2)$$ which is obviously true.

Now we assume the initial statement is true and wish to show that this holds for $n + 1$ and we are done. In other words, show

$$\sum_{j = 1}^{n + 1}j(j+1) = \dfrac{1}{3}(n+1)((n+1)+1)((n+1)+2)$$ $$\sum_{j = 1}^{n + 1}j(j+1) = \dfrac{1}{3}(n+1)(n+2)(n+3)$$

It's obvious that the extended sum is the initial sum plus the added term:

$$\sum_{j = 1}^{n + 1}j(j+1) = \sum_{j = 1}^{n}j(j+1) + \sum_{j = n + 1}^{n+1}j(j+1)$$

And we are assuming the initial sum to be true as part of our induction argument. Therefore, by substitution: $$\sum_{j = 1}^{n + 1}j(j+1) = \dfrac{1}{3}(n)(n+1)(n+2) + \sum_{j = n + 1}^{n+1}j(j+1)$$ $$\sum_{j = 1}^{n + 1}j(j+1) = \dfrac{1}{3}(n)(n+1)(n+2) + (n+1)((n+1)+1)$$ $$\sum_{j = 1}^{n + 1}j(j+1) = \dfrac{1}{3}(n)(n+1)(n+2) + (n+1)(n+2)$$

If you expand the right hand side of what you wish to show and multiply this final line all the way out, you will see the connection. Factoring at the end will not help you. For a formal proof just work your way down one side and back up the other.

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Observe that $$\sum\limits_{j=1}^{n} j^2 = \frac {1} {6} n(n+1) (2n+1)$$ and $$\sum\limits_{j=1}^{n} j = \frac {n(n+1)} {2}.$$

EDIT $:$

To prove the first sum consider the equality $(t+1)^3 - t^3 = 3t^2 + 3t +1.$ Take $t=1,2, \cdots, n.$ Then we get by summing the equality for $t=1,2, \cdots,n$ $$3\sum\limits_{t=1}^{n} t^2 = \sum\limits_{t=1}^{n} ((t+1)^3 - t^3) - 3 \left ( \sum\limits_{t=1}^{n} t \right ) - n.$$

Can you proceed now?

What is $\sum\limits_{t=1}^{n} ((t+1)^3 - t^3)$?

Observe that $\sum\limits_{t=1}^{n} ((t+1)^3 - t^3) = (n+1)^3 -1.$ Also you know that $\sum\limits_{t=1}^{n} t = \frac {n(n+1)} {2}.$ Now you should plug all these things to get the required answer.

So $$\sum\limits_{t=1}^{n}t^2 = \frac 1 3 \left ((n+1)^3 - 1 - \frac {3n(n+1)} {2} - n \right ).$$

I leave the further simplifications to you.

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  • $\begingroup$ I just want to see how it is concluded if it can be proven or not. $\endgroup$ – Chris Michael Apr 9 at 18:17
  • $\begingroup$ By the way can you see it now @Chris Michael? $\endgroup$ – Dbchatto67 Apr 9 at 18:17
  • $\begingroup$ Which one has to be proved?? $\endgroup$ – Dbchatto67 Apr 9 at 18:18
  • $\begingroup$ I'm really learning about this topic, and I wanted to take that exercise as a reference $\endgroup$ – Chris Michael Apr 9 at 18:18
  • $\begingroup$ Do you know the above two well known sums mentioned in my answer? $\endgroup$ – Dbchatto67 Apr 9 at 18:19

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