# if n is a positive integer, prove that, 1 cdot 2+2 cdot 3+3 cdot 4+⋯+n cdot (n+1) =(n(n+1)(n+2))/(3)

Discrete math induction problem.
if n is a positive integer, prove that,
$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot \left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}$
Basic step is 1, then assume k is true and we get,
$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +k\cdot \left(k+1\right)=\frac{k\left(k+1\right)\left(k+2\right)}{3}$
After that I am stuck because the answer key is telling me that the next step is to add $\left(k+1\right)\left(k+2\right)$, but I don't know why we do that. Aren't we supposed to just plugin $k+1$ into all k variables. Why are we adding $\left(k+1\right)\left(k+2\right)$ to both sides?
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Bianca Chung
Step 1
So we want to prove that
$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\left(n+1\right)=\sum _{i=1}^{n}i\left(i+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\mathrm{\forall }n\in \mathbb{N}$
The base case $n=1$:
Note that $1\cdot 2=2$ and that $\frac{1\cdot 2\cdot 3}{3}=2$. Thus the base case holds.
Suppose that $\mathrm{\exists }k\in \mathbb{N}$ such that
$\sum _{i=1}^{k}i\left(i+1\right)=\frac{k\left(k+1\right)\left(k+2\right)}{3}$. Now consider the case that $n=k+1$
$\sum _{i=1}^{k+1}i\left(i+1\right)=\left(n+1\right)\left(n+2\right)+\sum _{i=1}^{k}i\left(i+1\right)$
Step 2
By the induction hypothesis (we claimed that for $n=k$ the formula held) we have:
$\sum _{i=1}^{k+1}i\left(i+1\right)=\left(k+1\right)\left(k+2\right)+\frac{k\left(k+1\right)\left(k+2\right)}{3}=\frac{{k}^{3}+6{k}^{2}+11k+6}{3}=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{3}$
Therefore, by induction, the formula is true $\mathrm{\forall }n\in \mathbb{N}$
###### Did you like this example?
Cristofer Graves
Step 1
Based on hypothesis:
$\sum _{i=1}^{k}i\left(i+1\right)=\frac{1}{3}k\left(k+1\right)\left(k+2\right)$
it must be shown that:
$\sum _{i=1}^{k+1}i\left(i+1\right)=\frac{1}{3}\left(k+1\right)\left(k+2\right)\left(k+3\right)$
Here $k+1$ is 'plugged in' for k.
Step 2
That is the inductionstep. How to do that? Well,
$\sum _{i=1}^{k+1}i\left(i+1\right)=\sum _{i=1}^{k}i\left(i+1\right)+\left(k+1\right)\left(k+2\right)=\frac{1}{3}k\left(k+1\right)\left(k+2\right)+\left(k+1\right)\left(k+2\right)$
(in the last equality you are using the hypothesis)
So if you can show that the RHS of this indeed equals:
$\frac{1}{3}\left(k+1\right)\left(k+2\right)\left(k+3\right)$
then you are ready. Working out the RHS comes to adding $\left(k+1\right)\left(k+2\right)$ to $\frac{1}{3}k\left(k+1\right)\left(k+2\right)$