If f x -1- x n- then the value of f 0- f -0- f -0-2 - - f n 0- n - isA- nB- 2 nC- 2 n -1D- None of these

The correct option is B 2^nf'(x) = n(1 + x)^n 1, f"(x) = n(n 1)(1 + x)^n 2 f^n (x) = n!, f^n (0) = n! ⇒ 1 + n + n(n 1)/2! + ....+n!/n! =^nC_0 + ^nC_1 + ^nC ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If f x =1+ x ...

Question

If $f (x) = (1 + x)^{n}$ , then the value of $f (0) + f^{'} (0) + \frac{f^{''} (0)}{2!} + . . . . . + \frac{f^{n} (0)}{n!}$ is

n

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2^{n}

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2^{n - 1}

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N o n e o f t h e s e

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Solution

The correct option is B $2^{n}$
$f^{'} (x) = n (1 + x)^{n - 1}, f " (x) = n (n - 1) (1 + x)^{n - 2}$
$f^{n} (x) = n!, f^{n} (0) = n!$
$\Rightarrow 1 + n + \frac{n (n - 1)}{2!} + . . . . + \frac{n!}{n!}$
$=^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . . . +^{n} C_{n} = 2^{n}$

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Similar questions

Q. If

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

, then the value of

f (0) + f^{'} (0) + \frac{f " (0)}{2!} + . . . . + \frac{f^{n} (0)}{n!}

Q. If

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

, then the value of

f (0) + f^{'} (0) + \frac{f^{''} (0)}{2!} + \dots + \frac{f^{n} (0)}{n!}

Q. If

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

, then the value of

f (0) + f^{^{'}} (0) + \frac{f^{^{''}} (0)}{2!} + \dots \dots + \frac{f^{n} (0)}{n!}

Q. If

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

, then the value of

f (0) + f^{'} (0) + \frac{f^{''} (0)}{2!} + \dots + \frac{f^{n} (0)}{n!}

If $f (x) = (1 + x)^{n}$ then the value of $f (0) + f^{'} (0) + . . . + \frac{f^{n} (0)}{n!}$ is

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If f(x)=(1+x)n, then the value of f(0)+f′(0)+f′′(0)2!+.....+fn(0)n! is

The correct option is B 2nf′(x)=n(1+x)n−1,f"(x)=n(n−1)(1+x)n−2 fn(x)=n!,fn(0)=n! ⇒1+n+n(n−1)2!+....+n!n! =nC0+nC1+nC2+.....+nCn=2n

If $f (x) = (1 + x)^{n}$ , then the value of $f (0) + f^{'} (0) + \frac{f^{''} (0)}{2!} + . . . . . + \frac{f^{n} (0)}{n!}$ is

The correct option is B $2^{n}$
$f^{'} (x) = n (1 + x)^{n - 1}, f " (x) = n (n - 1) (1 + x)^{n - 2}$
$f^{n} (x) = n!, f^{n} (0) = n!$
$\Rightarrow 1 + n + \frac{n (n - 1)}{2!} + . . . . + \frac{n!}{n!}$
$=^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . . . +^{n} C_{n} = 2^{n}$