If f(x)=(2011 + x)^(n), where x is a rea...

If `f(x)=(2011 + x)^(n)`, where x is a real variable and n is a positive interger, then value of `f(0)+f'(0)+ (f'' (0))/(2!)+...+ (f^((n-1))(0))/((n-1)!)` is `-f(0)+f'(0)+ (f'' (0))/(2!)+...+ (f^((n-1))(0))/((n-1)!)` is -

A

`(2011)^(n)`

B

`(2012)^(n)`

C

`(2012)^(n)-1`

D

`n(2011)^(n)`

Verified by Experts

The correct Answer is:

C

`(2011)^(n)+.^(n)C_(1)(2011)^(n-1)+ .^(n)C_(2)(2011)^(n-1)+...+.^(n)C_(n-1)2011+.^(n)C_(n-1)`
`=(2011+1)^(n)-1`

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