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

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(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)}{

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

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(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

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