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

The correct option is D 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 ...

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 D $2^{n}$
$f^{'} (x) = n (1 + x)^{n - 1}, f ” (x) = n (n - 1) (1 + x)^{n - 2}$
$f^{n} (x) = n!, f^{n} (0) = n!$
$\Rightarrow 1 + n + \frac{n (n - 1)}{2!} + . . . . + \frac{n!}{n!} =^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . +^{n} C_{n} = 2^{n}$

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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. I: lf

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

, then the value of

f (0) + f^{'} (0) + \frac{1}{2!} f^{''} (0) + \dots + \frac{1}{n!}

f^{n} (0)

2^{n}

II:

f (x) = x^{n} + 4

, then the value of

f (1) + f^{^{'}} (1) + \frac{1}{2!} f^{^{''}} (1) + \dots + \frac{1}{n!} f^{n} (1)

2^{n} + 4

.
Which of the following is correct?

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+xn), then the value of f(0)+f′(0)+f"(0)2!+....+fn(0)n! is

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