If f(x)=1+nx+frac{n(n-1)}{2}x^2+frac{n(n-1)(n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,$f(x)=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\ldots+x^n$. By the Binomial Theorem,we know that $(1+x)^n = 1 + nx + \frac{n(n-1)}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given,$f(x)=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\ldots+x^n$. By the Binomial Theorem,we know that $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots + x^n$. Comparing this with the given expression,we have $f(x) = (1+x)^n$. Now,we find the first derivative of $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(1+x)^n = n(1+x)^{n-1}$. Next,we find the second derivative of $f(x)$: $f''(x) = \frac{d}{dx}[n(1+x)^{n-1}] = n(n-1)(1+x)^{n-2}$. Finally,substituting $x=1$ into the second derivative: $f''(1) = n(n-1)(1+1)^{n-2} = n(n-1)2^{n-2}$.

If $f(x)=1+nx+\frac{n(n-1)}{2}x^2+\frac{n(n-1)(n-2)}{6}x^3+\ldots+x^n$,then $f''(1)$ is equal to

Similar Questions

If $y = \left( \frac{ax + b}{cx + d} \right)^2$, then $2 \frac{dy}{dx} \frac{d^3y}{dx^3}$ is equal to

Let $y = \log_8 \left( \frac{1-x^2}{1+x^2} \right)$ for $-1 < x < 1$. Then at $x = \frac{1}{2}$,the value of $225(y' - y'')$ is equal to:

If $f(x) = 10 \cos x + (13 + 2x) \sin x$, then $f''(x) + f(x)$ is equal to

If $f$ is a polynomial,then the second derivative of $f(e^x)$ is

If $\frac{d^n y}{d x^n}=y_n$ and $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$,then $4 x y_2+2 y_1=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module