Sum of Cubes - Odd Integers

Here's the first part :

1^3   + 2^3  +  3^3   +  .....+  n^3   =  n^2 (n + 1)^2  / 4

Show that it's true for  n  = 1

1^3   =   (1)^2 (1 + 1)^2 / 4   =  4 / 4  =  1   ⇒   true

Assume that it;s true for n  = k

So  

1^3 + 2^3  + 3^3  +  .... +   k^3   =   k^2 (k + 1)^2 / 4

Show that it's true for k + 1

That is

1^3   +  2^3  + 3^3 +  .....+ k^3  + (k + 1)^3   =  [k+1]^2 [ (k +1) + 1) ]^2 / 4 

So we have

1^3   +  2^3  + 3^3 +  .....+ k^3  + (k + 1)^3   =  k^2 ( k + 1)^2 / 4  + (k + 1)^3

= [  k^2 (k + 1)^2  + 4(k + 1)^3] / 4

=  [ (k + 1)^2 ( k^2 + 4(k + 1) ]  / 4

= [ (k + 1)^2 ( k^2 + 4k + 4) ] /4

= [ (k + 1)^2 ( k + 2)^2 / 4 = 

= [ (k + 1)^2 [ (k + 1) + 1 ]^2 / 4

Which is what we wanted to show

For the second part....we can evaluate these sum-of-diferences  [ although there are probably faster ways, maybe  ??? ]

1         28         153          496          1225              2556

     27         125         343         729              1331

             98           218       386           602

                   120          168       216

                            48           48

We will generate a 4th power polynomial.....which is consistent with your second formula

Sorry... got to go to bed, now  !!!   ....LOL!!!

HI Cphill

I did not really understand exactly second part of question but I have interpreted it as follows:

Define Pattern :

S_{c = 1^{3 +} 3^{3 +} 5³..........+ (2n-1)³  =  n²(2n²-1)  (as per formula for sum of cubes for odd integers)}

Now Sum of cubes for n+(n+1)

S_{2n+1 = n²(2n²-1) + [ 2(n+1)-1]³}

= 2n⁴- n² + [2n+1]³

= 2n⁴- n²  + 8n³+ 12n² + 6n + 1

= (2n⁴ + 4n³ + 2n²) + (4n³ + 8n² + 4n) + (n² + 2n + 1)

= 2n²(n+1)^{2 +} 4n(n+1)² + (n+1)²

⁼ (n+1)²(2n² + 4n + 1)