codeBuilder

asked on

# 1/(1-x)^k

I didn't understand the circled part in the attached photo ?

Is it binomial ? However, -k may be negative and there is no negative number in binomial theory ?

question.png

Is it binomial ? However, -k may be negative and there is no negative number in binomial theory ?

question.png

yes it is binomial, k is a positive integer.

Wiki has some information about negative binomials

http://en.wikipedia.org/wiki/Binomial_coefficient

Wiki has some information about negative binomials

http://en.wikipedia.org/wiki/Binomial_coefficient

if Bin(k,n) = k!/(n!(k-n)!

then

Bin(-k,n) =(-1)^n *k!/(n!(k-n)!

= (-1)^n *Bin(k,n)

So for instance

1/(1-x) = 1+ x + x² + x³ + ....

To really understand why this is the case requires going beyond Pascal's triangle. These ideas are often presented without sufficient justification, in essence they are saying:- believe me it is true and it works.

The coefficients for -ve and fractional powers come from looking at the problem as a Taylor Series expansion.

then

Bin(-k,n) =(-1)^n *k!/(n!(k-n)!

= (-1)^n *Bin(k,n)

So for instance

1/(1-x) = 1+ x + x² + x³ + ....

To really understand why this is the case requires going beyond Pascal's triangle. These ideas are often presented without sufficient justification, in essence they are saying:- believe me it is true and it works.

The coefficients for -ve and fractional powers come from looking at the problem as a Taylor Series expansion.

ASKER

Bin(k,n) = k!/(n!(k-n)!

Bin(-k,n) =(-1)^n *k!/(n!(k-n)!

Each has one paranthesis which don't have any mathcing paranthesis?

Can you fix it please @GwynforWeb ?

Bin(-k,n) =(-1)^n *k!/(n!(k-n)!

Each has one paranthesis which don't have any mathcing paranthesis?

Can you fix it please @GwynforWeb ?

ASKER CERTIFIED SOLUTION

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Look here:

< http://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_to_negative_integers >

< http://mathworld.wolfram.com/BinomialCoefficient.html >