`sum_(r=0)^n(-1)^r .^n C_r[1/(2^r)+3/(2^(2r))+7/(2^(3r))+(15)/(2^(4r))+ .....mt e r m s]= (2^(m n)-1)/(2^(m n)(2^n-1))`

Explanation

`because (1-x)^(4)= sum_(r=1)^(n) (-1)^(r)""^(n)C_(r) x^(r)`
Let `P = sum_(r=1)^(n) (-1)^(r)""^(n)C_(r){((1)/(2))^(r) + ((3)/(4))^(r) + ((7)/(8))^(r) + ((15)/(16))^(r) + ..."upto m terms"}`
`=sum_(r=1)^(n) (-1)^(r)""^(n)C_(r-1). ((1)/(2))^(r) + sum_(r=1)^(n) (-1)^(r)""^(n)C_(r). ((3)/(4))^(r) + sum_(r=1)^(n) (-1)^(r)""^(n)C_(r) ((7)/(8))^(r) + sum_(r=1)^(n) (-1)^(r)""^(n)C_(r) ((15)/(16))^(r) + ..."upto m terms"`
`= (1-(1)/(2))^(n)+ (1-(3)/(4))^(n) + (1- (7)/(8))^(n)+ (1-(15)/(16))^(n)+ ..."upto m terms"`
`= ((1)/(2))^(n)+ ((1)/(2))^(2n) + ( (1)/(2))^(3n)+ ((1)/(2))^(4n)+ ..."upto m terms"`
`= (((1)/(2))^(n)[1-{((1)/(2))^(n)}^(m)])/(1-((1)/(2))^(n)) `
` = ((2^(mn)-1))/(2^(mn) (2^(n) -1))` .