Question

If $f\left(x\right)=(2011+x{)}^n$, where $x$ is a real variable and $n$ is a positive integer, then the value of
$f\left(0\right)+f^{\prime }\left(0\right)+\frac{f"\left(0\right)}{2!}+....+\frac{f^{(n-1)}\left(0\right)}{(n-1)!}$ is.

• A. $(2011{)}^n$
• B. $(2012{)}^n$
• C. $(2012{)}^n-1$
• D. $n(2011{)}^n$

Answer 1

The correct option is C $(2012{)}^n-1$

$f\left(x\right)={(2011+x)}^{n}f\left(0\right)+f^{{}^{\prime }}\left(0\right)+\frac{f^{{}^{\prime \prime }}\left(0\right)}{2!}+\frac{f^{{}^{\prime \prime \prime }}\left(0\right)}{3!}+.........\frac{f^{n-1}\left(0\right)}{(n-1)!}=2011+\frac{n{\left(2011\right)}^{n-1}}{2!}+\frac{n(n-1){\left(2011\right)}^{n-2}}{3!}+......\frac{\left(n\right(n-1)....2!)\left(2011\right)}{(n-1)!}=C_0^n\left(2011\right)+C_1^n{\left(2011\right)}^{n-1}+......C_{n-1}^n\left(2011\right)+C_n^n{\left(2011\right)}^0-1$

$={(2011+1)}^n-1$

$={\left(2012\right)}^n-1$

Hence, option $C$ is correct