If `f(x)=x^(n)`, then the value of `f(1)-(f'(1))/(1!)+(f''(1))/(2!)-(f'''(1))/(3!)+(f''''(1))/(4!)-....+((-1)^(n)f^(n)(1))/(n!)` is

Options

A. 1
B. `2^(n)`
C. `2^(n-1)`
D. 0

Answer

D

Explanation

We have,
`f(x)=x^(n)`
`f^(r)(x)=n(n-1)(n-2)......(n-(r-1))x^(n-r)`
`implies" "f^(r)(x)=(n!)/((n-r)!)x^(n-r)impliesf^(r)(1)=(n!)/((n-r)!)`
`:." "f(1)-(f'(1))/(1!)+(f''(1))/(2!)-(f'''(1))/(3!)+...+((-1)^(n)f^(n)(1))/(n!)`
`=underset(r=0)overset(n)sum(-1)^(r)(f^(r)(1))/(r!)," where"f^(0)(1)=f(1)`
`=underset(r=0)overset(n)sum(-1)^(r)(n!)/((n-r)!r!)=underset(r=0)overset(n)sum(-1)^(r)""^(n)C_(r)=0.`