Re: Pattern between cubes

On Mar 18, 3:17 pm, "Divij Rao" <divij_urdb...@xxxxxxxxxxx> wrote:
On Mar 18, 11:31 am, jiahao_anti-addictga...@xxxxxxxxxxx wrote:



On Mar 18, 1:26 pm, "Divij Rao" <divij_urdb...@xxxxxxxxxxx> wrote:

On Mar 18, 12:36 am, Gottfried Helms <h...@xxxxxxxxxxxxx> wrote:

Am 17.03.2007 17:27 schrieb Divij Rao:> the ans is: [n(n+1)/2]^2
hope its correct if u want its proof, its quite lengthy, but if u
desire, let me know, i will type it/post it.

Well, now the next degree... :-)

sum of fourth powers...

and then, seach a pattern to generalize -you'll
possibly will have as much fun as Hans Faulhaber and
Jacob Bernoulli ...

sum of fourth power is?

Gottfried Helms

is it correct?

Yes and I have doubts on whether you generalize the formula using the
pattern , Rao .
I am working one a formula that can find the sums of all powers , just
by using variables .- Hide quoted text -

- Show quoted text -

its not pattern that i foll0w always, there are ways i can derive
formulae

for ex: sum of cubes
observe the identity,
[x(x+1)]^2 - [x(x-1)]^2=4x^3

let f(x)=4x^3,
jus put x=1,2,3,...n, without simplifying, add the eqns, we getall
terms cancelled,
we have, 4summationx^3=[n(n+1)]^2,
1^3+2^3+...n^3=[n(n+1)]^2

we can use suitable identiteis to derive such formulae. similarly for
x^4,5,6,...

regards,
Divij

How about squares ? Total sum of squares . It should be easy , I got
my formula written down already , what's yours ?

.

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