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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. ${}{}$

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!!!

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What exactly are you trying to prove. You have put that you want to prove $1 \times 2 + 2 \times 3^2 \times 4 + (n-1)n = \frac{1}{3}n(n^-1).$ How does the LHS depend on $n$, it is not clear. I think you mean to have some kind of sum, perhaps $1\times 2 + 2\times 3 + 3\times 4 + \cdots + (n-1)n.$ – Fly by Night Dec 11 '12 at 17:14
He means $1*2+2*3+3*4+4*5+(n-1)*n$ but he wrote a "*" in stead of a "+" I guess – barto Dec 11 '12 at 17:17
shouldnt it be $1*2+2*3+3*4+...+(n-1)*n$? – CBenni Dec 11 '12 at 17:17
up vote 3 down vote accepted

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.

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$n(n+1)$ 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

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



and $a_n=n(n+1)$ and the result holds for $n=1$, the assertion follows.

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ok so I get this from the $1*2+2*3+3*4 ect.$ – Le Chifre Dec 11 '12 at 17:19
Note that OP seems to be adding $i(i-1)$, not $i(i+1)$ – Ross Millikan Dec 11 '12 at 17:27
@RossMillikan It is just the same. He starts at $i=2$, I start at $i=1$. – Pedro Tamaroff Dec 11 '12 at 17:39
You are right, I missed it. I think it is better to stay with OP's limits or to show where the change is made explicitly. – Ross Millikan Dec 11 '12 at 17:41
@RossMillikan Hmm, probably. – Pedro Tamaroff Dec 11 '12 at 17:59

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 $$ \sum_{k=1}^{n+1} k(k+1) =\sum_{k=1}^{n} k(k+1) +(n+1)(n+2)=\frac{n(n+1)(n+2)}{3} +(n+1)(n+2)...(\text{easy algebra})...= \frac{(n+1)(n+2)(n+3)}{3} $$

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