JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

f(x)=x+\frac{x^2}{1 !}+\frac{x^3}{2 !}+\ldots+\frac{x^n}{(n-1) !}

, then

f(0)+f^{\prime}(0)+f^{\prime \prime}(0)+\ldots+f^n(0)

is equal to

View answer and explanation

Correct option: A

\frac{n(n+1)}{2}

Option A (

\frac{n(n+1)}{2}

) is correct because the sum of the derivatives evaluated at 0 forms an arithmetic progression.

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