prove that `sum_(r=0)^n(-1)^r^n C_r`. [`1/(2^r)+(3^r)/(2^(2r))+(7^r)/(2^(3r))+(15^r)/(2^(4r))+ `......up to m terms`]`= `(2^(m n)-1)/(2^(m n)(2^n-1))`

Explanation

`underset(r=0)overset(n)sum(-1)^(r)xx.^(n)C_(r)(1/(2^(r))+(3/4)^(r)+(7/8)^(r)+".....m terms")`
`= underset(r=0)overset(n)sum.^(n)C_(r)((-1)/(2))^(r)+underset(r=0)overset(n)sum.^(n)C_(r)((-3)/(9))^(n)+underset(r=0)overset(n)sum.^(n)C_(r)((-7)/(8))^(r)+".......m terms"`.
`=(1-(1)/(2))^(n) + (1-3/4)^(n) + (1-7/8)^(n)+(1-15/16)^(n)+"....."m` terms a
`= ((1)/(2^(n))[1-(1/(2^(n)))^(m)])/(1-1/(2^(n)))= (2^(mn)-1)/(2^(mn)(2^(n)-1))`