JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

1 of 1

Application +4 | -1

f(x)=(1+x)^{n}

, then the value of

f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\cdots

+\frac{1}{n !} f^{n}(0)

View answer and explanation

Correct option: B

2^{n}

The correct answer is B (

2^{n}

) because the given expression is the binomial expansion of

(1+1)^n

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