Jim sent in this solution, using the ideas from our hints.


a(n)=1+2+¼+n= n(n+1)
2



b(n)=12+22+¼+n2= n(n+1)(2n+1)
6



c(n)=13+23+¼+n3= n2(n+1)2
4


It's obvious that c=a2, from this.

Also, 2a=n2+n, so, solving the quadratic (and using the fact that n > 0), we get
n=
-1+   ____
Ö1+8a
 
2

.

Now substitute this for n in b, to get
b= n(n+1)(2n+1)
6
 = a
3
 ×(2n+1)= a
3
 ×   ______
Ö1+ 8a
 

. So
3b=a   ____
Ö1+8a
 

, so 9b2=a2+8a3.

Now we can combine these two expressions: 9b2=c+8cÖc, so 8cÖc = 9b2-c, so 64c3=81b4-18b2c+c2.

(It's easy to check that the expressions above in terms of n do work in this!)