Q33 / 90 · KVPY 2011 (SB/SX) · Mathematics

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

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

Solution

$(2011)^{\mathrm{n}}+{ }^{n} \mathrm{C}_{1}(2011)^{n-1}+{ }^{n} C_{2}(2011)^{n-2}+\ldots+{ }^{n} C_{n-1} 2011+{ }^{n} C_{n-1}$ $=(2011+1)^{\mathrm{n}}-1$

Hence, the correct answer is option: 'C'.

More Binomial Theorem Questions