Tom Mattson

OK, here is the solution in all the gory details.

P(k)=Σ_j=1^k=(1/4)k²(k+1)²

Expanding the rightmost side, I get:

P(k)=Σ_j=1³=(1/4)(k⁴+2k³+k²)

Now, look at P(k+1):

Σ_j=1^k+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:
Σ_j=1^k+1=Σ_j^kj³+(k+1)³

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

So, I have:

Σ_j=1^kj³+(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:

Σ_j=1^kj³+(k+1)³-Σ_j=1^kj³=(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