IT'S NOT FUCKING HOMEWORK IT'S JUST A SCREENSHOT FROM A BOOK

It's not going to have the same center, obviously

How do you prove Newton's binomial formula by induction without the use of combinatorics or pascal's triangle or any other such shit? You're cleary supposed to be able to do so.

>>8447711
http://math.stackexchange.com/questions/577240/fixed-point-in-plane-transformation

Here's the exercise.

As you can see it says

> Check that ANY composition of shift, rotation and homothety of a plane has a fixed point if k of homothety is less than 1

>>8447716
Assume [math](x+y)^n = \sum_{0 \le k \le n} {n \choose k} x^k y^{n-k}[/math]
Write [math](x+y)^{n+1} = (x+y)(x+y)^n[/math] and use the formula [math]{n \choose k} + {n \choose {k+1}} = {{n+1} \choose {k+1}}[/math]

>>8447723
Rewrite pls

>>8447729
you rewrite

>>8447731
Your tex doesn't work, you fucked it up.

File: 8447723.png (10KB, 582x117px) Image search: [Google]

10KB, 582x117px

>>8447733
u w0t m8 ?

>>8447739
> without the use of combinatorics or pascal's triangle or any other such shit

>>8447745
Um if you want to prove an identity involving binomial coefficients, it stands to reason that at some point you are going to have to use properties of the binomial coefficients.

File: Nbf.png (4KB, 468x43px) Image search: [Google]

4KB, 468x43px

>>8447750
There's SPECIFICALLY no combinatorics notation in the book so that retards know you can solve it with no combinatorics.

>>8447772
Ugh.. binomial notation is a *notation*.
That thing I wrote could be rewritten as [math]\frac{n(n-1)\dots n-k+1)}{k!} + \frac{n(n-1)\dots (n-k)}{(k+1)!} = \frac{(n+1)\dots (n-k+1)}{(k+1)!}[/math]. Just do exactly what I said and replace binomials by that disgusting expression with factorials and you'll get your "combinatorics-free" proof.

>>8447781
Nigger, the formula you use is from COMBINATORICS. And in the exercise there's specifically factorial notation, so you know you don't have to use combinatorics here.

>>8447784
Dude I don't know how I can explain it any further. There is no use of combinatorics in that formula above, it's a literal manipulation of fractions. Just use assume the formula g) is true, multiply it by a+b and then use the formula I wrote above.
I dunno if this is some kind of elaborate bait or something but I'm done explaining

>>8447785
You learn this formula in the combinatorics course HENCE it's combinatoric. You see now ffs?

>>8447785
Factorials are combinatorics since n! is defined as the number of permutation of a set with n elements.

>>8447786
>>8447796
Are you seriously retarded? Then don't use that notation and exand everything into products if numbers and do it manually so you don't get triggered.

>>8448001
Retard, it's not about notation - this FORMULA is a part of combinatorics course which is obviously something you can avoid using to prove the theorem.

>>8448027
>This formula is part of a cominatorics course
Nigger, that identify is something universally used. Grab the definition of n chose k and do the arithmetic involved. If you don lime that theb again just right it as n!/(n-k)!k! And try it directly. And factorial is just notation for a string of multiplications. You could have used the gamma function for all I care. The only other proof I can remember that isn't by induction uses REAL combibatorial principles not definitions and identities.

>>8448041
Fucking, sorry for the spelling. Me and my phone are retarded.