Come on, go ahead, prove that you're as smart as you

I can't figure out how to use a proof by induction on such a sum

>>7800025

>Capture d’écran

>frogposting

>>7800025
not sure what Pk(n) is supposed to be...

>>7800054
Maybe the k-1th polynomial, but out of which?

>>7800054
>>7800064
Pk-1(n) is a polynomial of degree at most k-1

>>7800025
Double induction on k and n.

>>7800068
LMAOing right now

>>7800068
>je viens de rentrer en prépa
>je me sens plus
>je viens caca-poster sur quatre chaîne

casse-toi grosse bouse, ton exo c'est de la merde qu'on donnerait même pas à un terminale par respect.

>>7800025
http://www.trans4mind.com/personal_development/mathematics/series/sumsBernoulliNumbers.htm

Je vais faire tes devoirs, jeune MPSI lissencéphale.
Tu veux calculer [math]\sum_{i \,=\, 1}^n i^k[/math]. Tu cherches un polynôme [math]P(X)[/math] tel que [math]P(X\,+\,1) \,-\, P(X) \,=\, X^k[/math] et tu en déduis par somme téléscopique :
[eqn]\sum_{i \,=\, 1}^n i^k \,=\, \sum_{i \,=\, 1}^n [P(i \,+\, 1) \,-\, P(i)] \,=\, P(n \,+\, 1) \,-\, P(1)[/eqn]
De rien, crétiniste illettré X-ENS wannabe sur le point de chialer parce qu’il s’est tapé un AO à l’épreuve de physique de Centrale.

>>7800025

>posting arbitrary calculation which is dependent on your educational level

Why dosen't OP prove he's intelligent and do it and while he's at it prove he didn't just read it in some book

>>7800000

>>7800204
>caca-poster sur quatre chaîne
kek

>>7800025
Oh awesome. I thought I was the first one who came up with this.
There's also another approach to this problem.
[math]f(x) = \sum_{n=1}^{x} j^{k} = \sum_{n=1}^{x} (x-j)(j^k - (j-1)^k )[/math]
And you can prove the terms on the right side can be rewritten to represent a summation that is expressable in terms of x. Which is pretty cool, because you can get rid of the summation function!

>>7800840
[math]f(x) = \sum_{j=1}^{x} j^{k} = \sum_{j=1}^{x} (x-j)(j^k - (j-1)^k )[/math] *

This is pretty cool. The coefficients of the polynomial can be found by laplacian interpolation on the first n numbers of the sequence, but I don't think that can be used for an existence proof in general.
I'll try to think about it now

>>7800025
You can prove pretty easily using generating functions that
[math]\sum_{k=1}^x k^n = \int_0^x (-1)^n \operatorname{B}(n,-t) dt =(-1)^{n+1} \frac{\operatorname{B}(n+1,-x)-\operatorname{B}(n+1)}{n+1}[/math]

So the [math]k^{\text{th}}[/math] coefficient is [math](-1)^{n+1-k} \binom{n+1}{k} \frac{\operatorname{B}(n+1-k)}{n+1}[/math].

For [math]k=n+1[/math] you get [math]\frac{\operatorname{B}(0)}{n+1} = \frac{1}{n+1}[/math].
For [math]k=n[/math] you get [math]-\binom{n+1}{n} \frac{\operatorname{B}(1)}{n+1} = -\operatorname{B}(1) = \frac{1}{2}[/math].