prove without induction (1 Viewer)

[Sirius Black]

Could anyone prove the following eqn by not using mathematical induction, but straight forward algebra?
1³+2³+...+n³=(1/4) n²(n+1)²

I've noticed that 1³+2³=(1+2)²
1³+2³+3³=(1+2+3)²
.
.
so in the end, the LHS=(1+2+3..+n)²=(1/4) n²(n+1)²

By strictly speaking, my work above isn't really a proof
could any1 help plz?

 

[Slidey]

1^3+2^3+...+n^3=(1/4)n^2(n+1)^2
Assume for n
Test for n+1:
LHS=(1/4)n^2(n+1)^2 + (4/4)(n+1)^3
=(n+1)^2(n^2+4n+4)/4
=(n+1)^2(n+2)^2/4
RHS=(1/4)(n+1)^2(n+2)^2=LHS

 

[Sirius Black]

lolz, I need the proof without using the induction...u didnt read the question

 

[Slidey]

It is my hallmark trait.

So you want to find what 1^3+2^3+...+n^3 equals?
The following is an obvious identity: (k+1)^4-k^4=4k^3+6k^2+4k+1
When k=1: 2^4-1^4 = 4.1^3+6.1^2+4.1+1
When k=1: 3^4-2^4 = 4.2^3+6.2^2+4.2+1
When k=1: 4^4-3^4 = 4.3^3+6.3^2+4.3+1
...
When k=n-1: n^4-(n-1)^4 = 4(n-1)^3+6(n-1)^2+4(n-1)+1
When k=n: (n+1)^4-n^4 = 4n^3+6n^2+4n+1
Add LHS and RHs downwards.
LHS=(n+1)^4-1
š means sigma
RHS= š (4k^3+6k^2+4k+1) from 1 to n
RHS= 4š(k^3) + 6n(n+1)(2n+1)/6+4n(n+1)/2+n
LHS=RHS
(n+1)^4-1=4š(k^3) + 6n(n+1)(2n+1)/6+4n(n+1)/2+n
4š(k^3)=(n+1)^4-1 - n(n+1)(2n+1) - 2n(n+1) - n
4š(k^3)=(n+1)^4 - n(n+1)(2n+1) - 2n(n+1) - (n+1)
4š(k^3)=(n+1)(n^3+3n^2+3n+1-2n^2-n-2n-1)
4š(k^3)=(n+1)(n^3+n^2)
š(k^3)=n^2(n+1)^2 /4

Of course, this assumes that you know the sum of each natural number squared is n(n+1)(2n+1)/6 - though it can be derived the same way as above so that's OK.

 

[want2beSMART]

errr what topic is this under...

ITS SO HARD *CRIES*

 

[Slidey]

Oh, well... I don't know if this is under any topic. It's just a method I know of finding summations for strange series.