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Solved

Posted on 2006-06-10

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.

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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4 Comments

Can you find the difference between the sum for n and the sum for n+1?

((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

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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