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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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    shouldnt it be $1*2+2*3+3*4+...+(n-1)*n$?2012-12-11

3 Answers 3

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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

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

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.

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    @RossMillikan Hmm, probably.2012-12-11
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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} $