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if f(x) = x^n then the value of f(1) - (...

if `f(x) = x^n` then the value of `f(1) - (f'(1))/(1!) + (f''(1))/(2!) + ---+((-1)^n f''^--n times (1))/(n!)`

A

`2^(n)`

B

`2^(n-1)`

C

0

D

1

Text Solution

Verified by Experts

The correct Answer is:
C
Given , ` f(x) = x^(n) rArr f(1) = 1 `
Now ` f'(x) = nx^(n-1) rArr f'(1) = n`
` f''(x) n(n-1)x^(n-2) rArr f''(1) = n(n-1)`
` f''underset(vdots)((x)) n(n-1)x^(n-2) rArr f''(1) = n(n-1)`
` f^(n) (x) = n(n-1)(n-2)....2*1`
` rArr f^(n) (1) = n(n-1)(n-2)....2*1`
Now ,`f(1) - (f'(1))/(1!) + (f''(1))/(2!) - (f'''(1))/(3!) + ...+ ((-1)^(n) f^(n) (1))/(n!)`
` = 1 (n)/(1!) + (n(n-1))/(2!)-(n(n-1)(n-2))/(3!)`
` + ...+ ((-1)^(n) n(n-1)(n-2)...1)/(n!) = (1-1)^(n) = 0 ` .
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