`sum_(r=0)^n (-1)^r .^nC_r (1+rln10)/(1+ln10^n)^r`

Options

A. 1
B. -1
C. n
D. none of these

Answer

d

Explanation

Let `log_(e) 10 = x` . Then
`sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r) (1 + r log_(e)10)/((1 + log_(e) 10^(n))^(r) ) `
`sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)(1 + rx)/((1 + nx)^(r))`
`sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)((1 )/(1 + nx))^(r)+sum_(r=0)^(n) (-1)^(r) (n)/(r) ""^(n-1)C_(r-1) (rx)/((1 + nx)^(r))`
`sum_(r=0)^(n) (-1)^(r) ""^(n)C_(r)((1 )/(1 + nx))^(r)- (nx)/(1 + nx)sum_(r=1)^(n) (-1)^(r-1) ""^(n-1)C_(r-1) (1)/((1 + nx)^(r))`
` = (1 - (1) /(1+ nx))^(n) - (nx)/(1 + nx) (1 - (1)/(1 + nx))^(n-1)`
`= ((nx)/(1+nx))^(n) = ((nx)/(1+nx))^(n) = 0` .