Sum of cubes - please verify my proof

[iainfs]

Homework Statement

Prove

$$\sum^{2n}_{r=1} r^3 = n^2(2n+1)^2$$

by mathematical induction.

My proof is below but it is rather long winded. Is there a quicker way of doing the algebraic manipulation? Preferably one that forgoes the polynomial division and multiplying out all the brackets!

The Attempt at a Solution

Base case, $$n=1$$:

$$\sum^{2 \times 1}_{r=1} r^3 = 1^2 (2 \times 1+1)^2 = 1 \times 9 = 9$$

$$= 1^3 + 2^3 = 1 + 8 = 9$$

Inductive step, assume true for $$n=k, \quad k \in \mathbb{N}$$

$$\sum^{2k}_{r=1} r^3 = k^2(2k+1)^2$$

Adding the next two terms, $$n=k+1$$

$$\sum^{2(k+1)}_{r=1} r^3 = k^2(2k+1)^2 + (2k+1)^3 + (2k+2)^3$$

Multiplying out and collecting terms
$$\sum^{2k+2}_{r=1} r^3 = 4k^4 + 20k^3 + 37k^2 + 30k + 9$$

By factor theorem, (k+1) is a factor.
By long-hand polynomial division

$$= (k+1)(4k^3 + 16k^2 + 21k + 9)$$

By factor theorem, (2k+3) is a factor.
By long-hand polynomial division

$$= (k+1)(2k+3)(2k^2 + 5k + 3)$$

Factorising the quadratic
$$= (k+1)(2k+3)(2k+3)(k+1)$$
$$= (k+1)^2(2k+3)^2$$
$$= (k+1)^2(2(k+1)+1)^2$$

So if the statement is true for $$n=k$$, it is true for $$n=k+1$$.
Therefore
$$\sum^{2n}_{r=1} r^3 = n^2(2n+1)^2$$
for all $$n \geq 1, \quad n \in \mathbb{N}$$

Thanks for any suggestions on making this simpler.

 

[lanedance]

Hi iainfs

how about this

$$\sum^{2(k+1)}_{r=1} r^3 = k^2(2k+1)^2 + (2k+1)^3 + (2k+2)^3$$

Then re-write everything in terms of k+1

$$\sum^{2(k+1)}_{r=1} r^3 = ((k+1)-1)^2(2(k+1)-1)^2 + (2(k+1)-1)^3 + (2(k+1))^3$$

then to make the algebra simpler change variable to m = k+1
$$\sum^{2(k+1)}_{r=1} r^3 =\sum^{2m}_{r=1} r^3 = (m-1)^2(2m-1)^2 + (2m-1)^3 + (2m)^3$$

and then group terms and see where we end up
$$\sum^{2m}_{r=1} r^3 = ((m-1)^2 + (2m-1))(2m-1)^2 + (2m)^3$$
$$= m^2(2m-1)^2 + (8m)m^2$$
$$= m^2((2m-1)^2 + 8m)$$
$$= m^2((4m^2 -4m +1) + 8m)$$
$$= m^2(4m^2 +4m +1)$$
$$= m^2(2m +1)^2$$
...sweet

I always find that variable change always makes it easier for me...

 

[iainfs]

Thank you! Knew there would be a more elegant way.

 

[hsarp71]

Doesn't this work for all 2n in N too? (not just n in N) for 2n>=1.
Because adding (2n+1)^3 to both sides gives,
Sum(from r=1 to 2n+1) of r^3
= n^2*(2n+1)^2+(2n+1)^3
=(2n+1)^2*(n+1)^2
=((2n+1)/2))^2+((2n+1)+1)^2
and the initial cases of 2n=1, 2n=2, 2n=3 all work.

 

[lanedance]

and the initial cases of 2n=1, 2n=2, 2n=3 all work.

n is an integer, so not too sure what you mean by 2n=3

 

[hsarp71]

no I'm saying that it works for all 2n in N. so n does not have to be an integer for this to work- n can be 0.5, 1, 1.5, 2,...