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Math Induction

Posted on 2006-06-10
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I am trying to figure out a problem and need some help.
The problem is listed below.

Using induction, verify that each equation is true for every positive integer n.
1 · 2 + 2 · 3 + 3 · 4 + ......+ n ( n + 1 ) =       (n (n + 1)(n + 2) ) / 3

I can get the number 1 to work for n but numbers above that I cannot. Any help would be appreciated.
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Question by:rcanter
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4 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 16879675
Assuming it works for n (e.g. 1) you need to prove it works for n+1
Can you find the difference between the sum for n and the sum for n+1?
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Author Comment

by:rcanter
ID: 16879718
I cannot get n+1 to work, although I may be doing something wrong. I do not understand induction much and this is the first time I have tried, although it has been a long night and day of trying :-(
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LVL 18

Assisted Solution

by:JR2003
JR2003 earned 200 total points
ID: 16880170
You need to show that
((n+1)(n + 2)(n + 3) ) / 3     is equal to     (n (n + 1)(n + 2) ) / 3 + ((n+1)(n+2))

=(n (n + 1)(n + 2) ) / 3 + (3(n+1)(n+2))/3

which simplifies to:
((n+1)(n + 2)(n + 3) ) / 3
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LVL 37

Accepted Solution

by:
Harisha M G earned 300 total points
ID: 16880258
The principle of mathematical induction is very simple. You show that it works for atleast one number (generally 1) and then prove that if it works for n, it works for n+1 too.

That way, if it works for 1, then it should work for 1+1 = 2 also.
Since it works for 2, it should work for 2+1 = 3 also, and so on..

You said you can find that it works for 1. So, I will show the other part...

1 · 2 + 2 · 3 + 3 · 4 + ......+ n ( n + 1 ) =      (n (n + 1)(n + 2) ) / 3

Now, you need to add the "next term of the last term" of the LHS to both the sides..
Last term of LHS = n(n+1)
Next term (Change n to n+1, that's all)  = (n+1)(n+1 + 1) = (n+1)(n+2)

So, adding it to both the sides,

1 · 2 + 2 · 3 + 3 · 4 + ......+ n(n+1) + (n+1)(n+2) =  (n(n+1)(n+2))/3 + (n+1)(n+2)

Which is nothing but..

1 · 2 + 2 · 3 + 3 · 4 + ...... + (n+1)(n+2) =  (n(n+1)(n+2))/3 + 3(n+1)(n+2)/3
                        = (n+1)(n+2)(n+3)/3

You can see that the RHS is nothing but the original RHS, but n being replaced by n+1.

That proves that if it works for n, it must work for n+1 too..
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