Induction prove, how to come to $n\cdot(n+1)$ I am trying to solve an induction problem. Here are the steps for the example.
Prove this equation $$
1\cdot2 + 2\cdot3 + 3\cdot 4 + 4\cdot 5+\dots + \cdots +(n-1)\cdot n  ={n\cdot(n-1)\cdot(n+1)\over 3 }
$$ for $n=2,3,4,5$ and prove that the equation is right for all natural numbers $n\ge 2$ with induction.
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induction beginning:
$$
\sum_{i=2}^n = {n\cdot(n-1)\cdot(n+1)\over 3} 
$$
-> this is clear to me!
induction hypothesis:
$$
\sum_{i=2}^n = {n\cdot(n+1)\cdot(n+2)\over 3} 
$$
->here you just put n+1, also clear to me
prove:
$$
\sum_{i=2}^n = {(n-1)\cdot n\cdot(n+1)\over 3} + n\cdot(n+1) = {n\cdot(n+1)\cdot(n+2)\over 3} 
$$ 
-> not clear
Then you put up the fuction to prove it, however I do not understand, why you add $n\cdot(n+1)$ and how to come to $n\cdot(n+1)$?
Thx for your answer!!!
 A: For the left side of your equations, you should show what is being added, as $$\sum_{i=2}^n(i-1)i$$
For your induction beginning, you just check for $n=2$, getting $1 \cdot 2=\frac {2 \cdot 1 \cdot 3}3=2$
Then the induction hypothesis is $$\sum_{i=2}^n (i-1)i= {n*(n-1)*(n+1)\over 3}$$
(note the difference from what you wrote)
Now you want to prove it for $n+1$ which would be $$\sum_{i=2}^{n+1} (i-1)i= {(n+1)*n*(n+2)\over 3}$$
So we have $$\sum_{i=2}^{n+1} (i-1)i=\sum_{i=2}^n (i-1)i+n(n+1)={n*(n-1)*(n+1)\over 3}+n(n+1)$$
The reason you add $n(n+1)$ is that is the new term when you raise the upper limit of the sum to $n+1$.  Your induction hypothesis shows you how to sum up to $n$, which you need to make use of.  Now see if you can work the right side to get the right side of your goal.
A: $n(n+1)$ is the next term you are adding. You're trying to prove that 
$$\sum_{i=1}^n i(i+1)=\frac{n(n+1)(n+2)}{3}$$
with the inductive hypothesis that 
$$\sum_{i=1}^{n-1} i(i+1)=\frac{n(n-1)(n+1)}{3}$$
Since
$$S_n+a_{n+1}=S_{n+1}$$
and $a_n=n(n+1)$
and the result holds for $n=1$, the assertion follows.
A: You want to prove $$\sum_{k=1}^{n} k(k+1) =\frac{n(n+1)(n+2)}{3}$$
induction beginning:
For $n=1$
$$
\sum_{k=1}^1 k(k+1)=2= {1\cdot(1+1)\cdot(1+2)\over 3} 
$$
holds
induction hypothesis:
Assume $$
\sum_{k=1}^{n} k(k+1) =\frac{n(n+1)(n+2)}{3}
$$
prove:
$$\sum_{k=1}^{n+1} k(k+1) =\frac{(n+1)(n+2)(n+3)}{3}$$
We have
$$
\sum_{k=1}^{n+1} k(k+1) =\sum_{k=1}^{n} k(k+1) +(n+1)(n+2)=\frac{n(n+1)(n+2)}{3} +(n+1)(n+2)...(\text{easy algebra})...=
 \frac{(n+1)(n+2)(n+3)}{3}
$$ 
