Is the series expansion of $\frac1{(1+x)^n}$ same as $\frac1{(1-x)^n}$ with $(-1)^r$

Question

The series expansion for $$\frac1{(1-x)^n} = \sum_{r=0}^{r=\infty}C_r^{|n|+r-1}x^r$$ Is the expansion of $$\frac1{(1+x)^n} = \sum_{r=0}^{r=\infty}(-1)^rC_r^{|n|+r-1}x^r$$ ?? (Where C is combination)

Answer 1

Yes you have the correct expansions with the exception of the absolute values on $n$. To see why we use the generalized binomial theorem: $$ (x+y)^s=\sum _{r=0}^{\infty }{s \choose r}x^{s-r}y^r, $$ which requires $|x|>|y|$ to converge if $s\notin\Bbb N_0$. Substituting $x=1$ and $s=-n$ one has $$ \frac{1}{(1+y)^n}=\sum _{r=0}^\infty {-n \choose r}y^r, $$ which converges for $|y|<1$. Now note $$ {-n \choose r}=(-1)^r\frac{(n)_r}{r!}=(-1)^r\frac{(n-1+r)!}{(n-1)!r!}=(-1)^r{n+r-1 \choose r}, $$ so that $$ \frac{1}{(1+y)^n}=\sum _{r=0}^\infty {n+r-1 \choose r}(-y)^r. $$ Finally substitute $y\mapsto -x$ and $y\mapsto x$ to get $$ \frac{1}{(1-x)^n}=\sum _{r=0}^\infty {n+r-1 \choose r}x^r $$ and $$ \frac{1}{(1+x)^n}=\sum _{r=0}^\infty (-1)^r{n+r-1 \choose r}x^r. $$