JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

f\left( x \right) = 1 + nx + \frac{{n\left( {n - 1} \right)}}{2}{x^2} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6}{x^3} + . \ldots + {x^n},

then,

f''\left( 1 \right)

is equal to

View answer and explanation

Correct option: C

n\left( {n - 1} \right){2^{n - 2}}

Option C (

n(n - 1)2^{n - 2}

) is correct because the given series is the binomial expansion of

(1 + x)^n

, and its second derivative evaluated at x=1 is

n(n - 1)2^{n - 2}

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