Binomial expansions for negative integers.

[Jadon]

Recently I was doing some revision of binomial theorem, a topic I took in extension 1 mathematics (HSC curriculum Australia) and came across an interesting question I haven't seen. The question was as follows.

Find the first 4 terms of the binomial (1-a)^-2. I thought upon this question and realized (1-a)^-2=1/(1-a)² , however this didn't help solve the question. I eventually realized I could use the binomial formula (x+y)ⁿ=xⁿ+nx^n-1y+(n(n-1))x^n-2y²/2+...+yⁿ substituting -n. However this doesn't really satisfy me, I can't understand how binomial expansions of negative integers seem to be infinite. I later came across a formula for this situation on wolfram
located here http://mathworld.wolfram.com/NegativeBinomialSeries.html This formula seems good but I can't seem to derive a proof myself. So my question would be can anybody show me a proof for this formula and also can anyone explain exactly how this type of binomial has infinite terms?

Thankyou.

 

[Ishuda]

Recently I was doing some revision of binomial theorem, a topic I took in extension 1 mathematics (HSC curriculum Australia) and came across an interesting question I haven't seen. The question was as follows.

Find the first 4 terms of the binomial (1-a)^-2. I thought upon this question and realized (1-a)^-2=1/(1-a)² , however this didn't help solve the question. I eventually realized I could use the binomial formula (x+y)ⁿ=xⁿ+nx^n-1y+(n(n-1))x^n-2y²/2+...+yⁿ substituting -n. However this doesn't really satisfy me, I can't understand how binomial expansions of negative integers seem to be infinite. I later came across a formula for this situation on wolfram
located here http://mathworld.wolfram.com/NegativeBinomialSeries.html This formula seems good but I can't seem to derive a proof myself. So my question would be can anybody show me a proof for this formula and also can anyone explain exactly how this type of binomial has infinite terms?

Thankyou.

There are several ways to do so. As for the infinite terms one might make the analogy of fractions that result in infinite terms, that is \(\displaystyle \frac{1}{\sqrt{2}}\, =\, 0.707....\); an infinite number of 'terms'.

One of the ways is to just do the long division

[FONT=courier new]           1 - 2a 
         ___________________
1+2a+a[SUP]2[/SUP]  | 1
           1+2a+a[SUP]2[/SUP]
          ---------
            -2a- a[SUP]2[/SUP]
            -2a-4a[SUP]2[/SUP]-2a[SUP]3[/SUP] 
            -----------
                3a[SUP]2[/SUP]+2a[SUP]3[/SUP] 
etc.[/FONT]

If you wanted to accept the initial wolfram formula but see how it is change into the second, consider the definition of the binomial coefficient:
_nC_k = n! / [ k! (n-k)! ] = \(\displaystyle \frac{n (n-1) (n-2) ... (n-k+1)}{k (k-1) (k-2) ... 1}\)
Now put in a -n instead and multiply by 1 in the form of (-1)^k * (-1)^k
_-nC_k = \(\displaystyle (-1)^k\, \frac{(-1)*(-n) (-1)*(-n-1) (-1)*(-n-2) ... (-1)*(-n-k+1)}{k (k-1) (k-2) ... 1}\)
to get
_-nC_k = \(\displaystyle (-1)^k\, \frac{n (n+1) (n+2) ... (n+k-1)}{k (k-1) (k-2) ... 1}\)
etc.