What is the expression for the series involving cubes and squares?

[aalmighty]

It's a nice li'l series. The question is to find an expression for the series


S=1^3-2^3+3^3-4^3+...-(2m-2)^3+(2m-1)^3

where m is a subset of N

I added and subtracted sum of even squares to get

S={1^3+2^3+3^3+4^3+...+(2m-1)^3}-{2(2^3+4^3+6^3+...+(2m-2)^3)}

Replaced the first part by [(2m-1)(2m)/2]^2 to get

S=[(2m-1)(2m)/2]^2-2(2^3+4^3+6^3+...+(2m-2)^3)

now All I need to complete the problem is to get the general formula for the sum of cubes of the first n even natural numbers, plug the values and evaluate. Can anyone please help?

Also, Please let me know if there is an alternative way to solve this problem.

 

[Tx]

I'd use the sum of an AP to solve this. Unless I've read the question wrong.

 

[HallsofIvy]

No, sum of cubes is not an arithmetic progression.

2³+ 4³+ 6³+ . . .
= 2³+ (2*2)³+ (2*3)³+ ...
= 2³(1+ 2³+ 3³+...)

Do you know the formula for the sum of cubes?

 

[aalmighty]

Oh! Thank you. I'm kicking myself for not figuring that out :)