The sum of the first n cubes
By Saul Foresta on Thursday, July 11, 2002 - 01:06 am:

I would appreciate help proving that-

13 +23 +...+n3 =[n(n+1)/2]2

I can do it using induction but i am interested in seeing a different proof.

By Brad Rodgers on Thursday, July 11, 2002 - 03:30 am:

We know that



n
å
k=0 
 xk=(1-xn)/(1-x)

.
Thus, differentiating and multiplying by x



n
å
k=1 
 k xk = x(xn+1-n xn - xn + 1)/(x-1)2

.
Repeat this two more times, and then let x tend to 1. You should be able to end up with the formula you've given above, but only with considerable work...


There's also a method using what are called Bernoulli numbers, but this is also fairly complex.

Perhaps someone else knows a simpler way, although I'd imagine induction is it on this particular problem,

Brad
By David Loeffler on Thursday, July 11, 2002 - 09:25 am:

You could try starting from n4 - (n-1)4 = 4n3 + (terms in n2 and lower powers). If you already have the summations for 12 + ... + n2 and so on, you can get the one for n3 . But induction is MUCH quicker.

David

By Julian Pulman on Thursday, July 11, 2002 - 06:52 pm:

How about this:


(x-1)4=x4-4x3+6x2-4x+1

Then (x-1)4-x4=4x3-6x2+4x-1.

Now, sum both sides from x=1 to x=n: I'll use the shorthand notation
å
= n
å
1 


n4=4 å
x3 - 6 å
x2+4 å
x-n


n4=4 å
x3-6(n(2n1)(n+1)/6)+4(n(n+1)/2)-n


Þ 4 å
x3=n(n3+2n2+n)=n2(n+1)2


Þ å
x3=n2(n+1)2/4

Of course, the formulas for n2, and n can be calculated recursively using the same method.
Julian

By Yatir Halevi on Thursday, July 11, 2002 - 08:31 pm:

Julian, your method is actually a famous method to find the sum of powers...

Yatir

By Arun Iyer on Thursday, July 11, 2002 - 09:40 pm:

This is another method to solve the problem.....

assume...
13 +23 +33 +....+n3 =A+Bn+Cn2 +Dn3 +En4 +Fn5 ........... ----- (1)

replace n by n+1,then
13 +23 +33 +....+n3 +(n+1)3 =A+B(n+1)+C(n+1)2 +D(n+1)3 +E(n+1)4 +F(n+1)5 ............. ------ (2)

then (2)-(1) gives,
(n+1)3 =B+C(2n+1)+D(3n2 +3n+1)+E(4n3 +6n2 +4n+1)+F(5n4 +....)+.............

This equation being true for all integral value of n,the coefficients of the respective powers of n on each side must be equal;thus F and all succeeding coefficients must be equal to zero..

therefore,
n3 +3n2 +3n+1=B+C(2n+1)+D(3n2 +3n+1)+E(4n3 +6n2 +4n+1)

Comparing the coefficients we get,
4E=1 .......... (I)
6E+3D=3 ....... (II)
4E+3D+2C=3 .... (III)
E+D+C+B=1 .... (IV)

Solving I,II,III and IV,we get
E=1/4
D=1/2
C=1/4
B=0

therefore,from 1,
13 +23 +33 +....+n3 =A+(n2 /4)+(n3 /2)+(n4 /4)

to determine A,put n=1 and we get A=0
therefore,
13 +23 +33 +....+n3 =(n2 /4)+(n3 /2)+(n4 /4)
rearrange the RHS and we get,
13 +23 +33 +....+n3 =[n(n+1)/2]2

hence proved...

love arun
P.S--> this question can also be solved by another method by finding the recurring relation...

By David Loeffler on Friday, July 12, 2002 - 09:02 am:

Julian, this is what I was suggesting in the third message in this thread.

The usefulness of this approach is that it provides an immediate inductive proof that all the f's exist 1k + ... + nk is a polynomial of degree k+1 in n. We can then just find the coeffs by solving simultaneous equations.

Incidentally, of course if the equations have a solution, this implies existence as well!

David

By Julian Pulman on Friday, July 12, 2002 - 09:41 am:

whoops, completely missed that message, sorry David!..I think it's quite a neat way to derive it though. I quite dislike proving things by induction because it requires you know an equation to begin with, I remember thinking earlier on this year how silly proving De Moivre by induction was, as De Moivre certainly wouldn't have know the formula before deriving it (I mean, it's not obvious is it? You'd think expanding the brackets would reveal a really complicated binomial expansion..)
I can't believe I entirely missed your post!

Julian