We've been given with following binomial expansion
$$(x+1)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3\cdots$$
How can I get the formula of $(1-x)^n$
[Math] Binomial expansion of $(1-x)^n$
binomial theorem
binomial theorem
We've been given with following binomial expansion
$$(x+1)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3\cdots$$
How can I get the formula of $(1-x)^n$
Best Answer
Well, as I understand it, we could write the binomial expansion as:
$$(1-x)^n= \sum_{k=0}^{n} \binom n k 1^{n-k}\,(-x)^k$$ $$\binom{n}{0}1^n (-x)^0 + \binom n 1 1^{n-1} (-x)+ \binom n 2 1^{n-2}(-x)^2 + \binom n 3 1^{n-3}(-x)^3 \ldots$$
which simplifies to $$1-nx+\frac{n(n-1)}{2!}\cdot x^2 -\frac{n(n-1)(n-2)}{3!} \cdot x^3 \ldots$$
Which is the answer everyone else has given.