Was just looking at binomial theorem, i am confused [help]

[seto6]

i was told the binomial theorem is as follows:

(1-x)^n = 1-nx+ (n(n-1)/2!)x^2 - (n(n-2)/2!)x^3 ...
not sure if this is right

could some one clear this doubt for me
any help is appreciated

was told this in a physics class

 

[tiny-tim]

Hi seto6!

(try using the X² and X₂ tags just above the Reply box )

It's usually written (A + B)ⁿ = ∑_i=1^{n n}C_i AⁱB^n-i

where ⁿC_i = n!/i!(n-i)!.

So in your case, (1 - x)ⁿ = ∑_i=1^{n n}C_i (-x)ⁱ

= 1-nx+ (n(n-1)/2!)x² - (n(n-1)(n-2)/3!)x³ ...

 

[HallsofIvy]

i was told the binomial theorem is as follows:

(1-x)^n = 1-nx+ (n(n-1)/2!)x^2 - (n(n-2)/2!)x^3 ...

This last term is wrong. It should be n!/(3!(n-3)!)= n(n-1)(n-2)/3!. In general the ith coefficient is n!/(i!(n- i)!)= n(n-1)(n-2)...(n-i+1)/i!

not sure if this is right

could some one clear this doubt for me
any help is appreciated

was told this in a physics class

Since your teacher is not here to defend himself, I am going to assume you miscopied.

 

[uart]

Yes that's (nearly) the best way to express the more general binomial theorem where "n" is not necessarily an integer.

Normally however you'd use "+" as the default (and just absorb the "-" into the "x" if you need negative). That is

(1 + x)^n = 1 + n x + n(n-1)/2! x^2 + n(n-1)(n-2)/3! x^3 + ...

For the case where n is a positive integer then the above series terminates at the (n+1)th term, when you get down to n(n-1)(n-2)…(n-n), as this and every following term is multiplied by zero. In this case it reduces to the more familiar binomial theorem as shown in the preceding two posts.

Also, while its convenient to have one of the terms fixed at 1 (especially for the non terminating series) its obviously not fully general in that form. Most general is (a+b)^n, but in that case I prefer to just factor out the a^n, as in a^n (1+ (b/a)^n), and proceed as before.

Hope that helps.