What is the Binomial Expansion of ${(1 + r)}^{- 1}$ $(1+r)^-1$ ?

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Answer 1 · 2015-07-18T07:16:39

If $| r | < 1$ $abs(r) < 1$ then
$(1+r)^(-1) = sum_(n=0)^oo(-1)^nr^n = 1 - r + r^2 - r^3 +...$

If $abs(r) > 1$ then
$(1+r)^-1 = sum_(n=0)^oo(-1)^nr^(-n-1) = 1/r-1/(r^2)+1/(r^3)-...$

If $abs(r) < 1$ then $sum_(n=0)^oo(-1)^nr^n$ converges and

$(1+r)sum_(n=0)^oo(-1)^nr^n=1 + r - r + r^2 - r^2 + r^3 - r^3 +... = 1$

So $(1+r)^-1 = 1/(1+r) = sum_(n=0)^oo(-1)^nr^n$

If $abs(r) > 1$ then $(1+r)^(-1) = 1/(1+r) = (1/r)/(1/r + 1) = (1/r)*(1+1/r)^(-1)$

and $abs(1/r) < 1$ , so we can use our previous formula to get:

$(1+r)^(-1) = (1/r)sum_(n=0)^oo(-1)^nr^(-n)=sum_(n=0)^oo(-1)^nr^(-n-1)$

That just leaves the case $abs(r) = 1$ , i.e. $r = +-1$

$(1+1)^-1 = 1/2$

$(1-1)^-1 = 1/0$ is undefined

What is the Binomial Expansion of (1+r)^-1(1+r)−1(1+r)^-1?