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

Question

4680 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-18T07:16:39

If #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