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Tom Mattson
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#6
Sep23-03, 09:25 PM
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OK, here is the solution in all the gory details.

P(k)=Σj=1k=(1/4)k2(k+1)2

Expanding the rightmost side, I get:

P(k)=Σj=13=(1/4)(k4+2k3+k2)

Now, look at P(k+1):

Σj=1k+1j3=(1/4)(k+1)2(k+2)2

I am going to manipulate both sides of this statement, one at a time:

Isolate (k+1)st Term On Left Side:
Σj=1k+1=Σjkj3+(k+1)3

Expand Right Side:
(1/4)(k4+6k3+13k2+12k+4)

So, I have:

Σj=1kj3+(k+1)3=(1/4)(k4+6k3+13k2+12k+4)

Now, remember P(k)? Assume it is true, and subtract it from P(k+1). It will be handy if you subtract the left side of P(k) from the left side of P(k+1), and do likewise with their right sides. I am going to denote both sides of P(k) with the color blue.

Then you get:

Σj=1kj3+(k+1)3-Σj=1kj3=(1/4)(k4+6k3+13k2+12k+4)-(1/4)(k4+2k3+k2)

And finally,

(k+1)3=k3+3k2+3k+1

which is an identity. Therefore, P(k)-->P(k+1).

edit: fixed superscript bracket