OK, here is the solution in all the gory details.
P(k)=Σ_{j=1}^{k}=(1/4)k^{2}(k+1)^{2}
Expanding the rightmost side, I get:
P(k)=Σ_{j=1}^{3}=(1/4)(k^{4}+2k^{3}+k^{2})
Now, look at P(k+1):
Σ_{j=1}^{k+1}j^{3}=(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}^{k}j^{3}+(k+1)^{3}
Expand Right Side:
(1/4)(k^{4}+6k^{3}+13k^{2}+12k+4)
So, I have:
Σ_{j=1}^{k}j^{3}+(k+1)^{3}=(1/4)(k^{4}+6k^{3}+13k^{2}+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}^{k}j^{3}+(k+1)^{3}Σ_{j=1}^{k}j^{3}=(1/4)(k^{4}+6k^{3}+13k^{2}+12k+4)(1/4)(k^{4}+2k^{3}+k^{2})
And finally,
(k+1)^{3}=k^{3}+3k^{2}+3k+1
which is an identity. Therefore, P(k)>P(k+1).
edit: fixed superscript bracket
