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The given summation is: $$\sum_{i=1}^n i(i+1)$$

The goal is to convert it into a formula which only uses n.

Solving this, I got the answer:

$$\frac{n}{3}(n+1)(n+2)$$

However, I don't believe the way I solved it is the intended method, despite arriving at the correct solution. This is because a hint is provided:

$$i(i+1) = \frac{(i+1)^3 - i^3 - 1}{3}$$

My solution did not benefit from this hint, thus I believe my method of solving was unintended.

Based on the hint, what method should be used to convert the summation into a formula?

C T
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    There's a typo in the hint: the fraction should be $;\dfrac{(i+\color{red}1)^3-i^3-1}3$. – Bernard Jan 29 '16 at 20:12
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    The hint suggests an obvious telescope structure. – Zhanxiong Jan 29 '16 at 20:12
  • Using the hint, the evaluation of the sum ultimately boils down to determining $\sum\limits_{i=1}^n \big[(i+1)^3-i^3\big]$, which can be easily seen to equal $$ \begin{array}{ccc} \color{#C00}{(1+1)}^3&-&1^3 \ \color{darkorange}{(2+1)}^3&-&\color{#C00}2^3 \ \color{green}{(3+1)}^3&-&\color{darkorange}3^3 \ &\vdots& \ \color{brown}{(n)}^3&-&\color{royalblue}{(n-1)}^3 \ (n+1)^3&-&\color{brown}n^3 \ \hline \text{What's left?} \end{array} $$ – Workaholic Jan 29 '16 at 20:17
  • BTW, what method did you use? It's probably valid and maybe even obliquely used the hint. – fleablood Jan 29 '16 at 20:20
  • The typo was corrected. Thanks for pointing that out.

    @fleablood: I just accepted the top-voted answer, and the second approach given there is the one I used. It struck me as being more obvious, given "i^2 + i", and I didn't see how changing it into that massive alternative helped.

    The answers really helped clarify things. Thanks everyone.

     – C T Jan 30 '16 at 16:41

3 Answers3

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For fixed $i$,

$\begin{eqnarray} \frac{(i+1)^3-i^3-1}{3}&=&\frac{i^3+3i^2+3i+1-i^3-1}{3}\\ &=&\frac{3i^2+3i}{3}\\ &=&i(i+1) \end{eqnarray}$

So

$\begin{eqnarray} \sum_{i=1}^ni(i+1)&=&\sum_{i=1}^n\frac{(i+1)^3-i^3-1}{3}\\ &=&\frac{1}{3}\sum_{i=1}^n[(i+1)^3-i^3-1]\\ &=&\frac{1}{3}\left(\sum_{i=1}^n(i+1)^3-\sum_{i=1}^ni^3-\sum_{i=1}^n1\right)\\ &=&\frac{1}{3}\left(\sum_{i=2}^{n+1}i^3-\sum_{i=2}^ni^3-1-\sum_{i=1}^n1\right)\\ &=&\frac{1}{3}((n+1)^3-1-n)\\ &=&\frac{1}{3}(n^3+3n^2+2n)\\ &=&\frac{n(n+1)(n+2)}{3} \end{eqnarray}$

Another approach is

$\sum_{i=1}^ni(i+1)=\sum_{i=1}^ni^2+i=\sum_{i=1}^ni^2+\sum_{i=1}^ni=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$ and simplifies.

sinbadh
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$\sum_{k=1}^n k(k+ 1) = \sum_{k=1}^n \dfrac{(i+1)^3-i^3-1}3=$

$1/3[\sum_{k=1}^n (i+1)^3- \sum_{k=1}^n i^3 - \sum_{k=1}^n1]=$

$1/3[\sum_{k=2}^{n+1} i^3- \sum_{k=1}^n i^3 - n]=$

$1/3[(n+1)^3- 1^3 - n]=$

$1/3[n^3 + 3n^2 + 3n - n]=$

$1/3[n^3 + 3n^2 + 2n]=$

$n/3[n^2 + 3n + 2]=$

$\dfrac{n(n+1)(n+2)} 3$

fleablood
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The hint is incorrect, as you can easily tell by epanding the RHS. They probably meant $$i(i+1)=\frac{(i+1)^3-i^3-1}{3}\ ,$$ and you can use this to evaluate the result as a telescoping sum: $$\eqalign{\sum_{i=1}^n i(i+1) &=\frac13\sum_{i=1}^n((i+1)^3-i^3-1)\cr &=\frac13((2^3+\cdots+n^3+(n+1)^3)-(1^3+2^3+\cdots+n^3)-(1+2+\cdots+n))\cr &=\frac13((n+1)^3-1-\frac12n(n+1))\cr}$$ and so on.

David
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