If f(x)=(2011+x)^n,where x is a real variable | vedclass

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(C) Given $f(x) = (2011+x)^n$. By Taylor's expansion of $f(x)$ about $x=0$,we have: $f(x) = f(0) + f^{\prime}(0)x + \frac{f^{\prime \prime}(0)}{2!}x^2

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$(2012)^n-1$

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(C) Given $f(x) = (2011+x)^n$. By Taylor's expansion of $f(x)$ about $x=0$,we have: $f(x) = f(0) + f^{\prime}(0)x + \frac{f^{\prime \prime}(0)}{2!}x^2 + \ldots + \frac{f^{(n)}(0)}{n!}x^n$. Since $f(x) = (2011+x)^n$,the binomial expansion is: $f(x) = \sum_{k=0}^{n} \binom{n}{k} (2011)^{n-k} x^k = (2011)^n + \binom{n}{1}(2011)^{n-1}x + \binom{n}{2}(2011)^{n-2}x^2 + \ldots + x^n$. Comparing the coefficients of $x^k$ in both expansions,we get $\frac{f^{(k)}(0)}{k!} = \binom{n}{k}(2011)^{n-k}$. The given expression is $S = f(0) + f^{\prime}(0) + \frac{f^{\prime \prime}(0)}{2!} + \ldots + \frac{f^{(n-1)}(0)}{(n-1)!}$. This is the sum of the first $n$ terms of the binomial expansion of $(2011+1)^n$ excluding the last term ($x^n$ term where $k=n$). $S = \sum_{k=0}^{n-1} \binom{n}{k} (2011)^{n-k} (1)^k$. We know that $(2011+1)^n = \sum_{k=0}^{n} \binom{n}{k} (2011)^{n-k} (1)^k = S + \binom{n}{n}(2011)^0(1)^n$. Therefore,$S = (2012)^n - 1$.

If $f(x)=(2011+x)^n$,where $x$ is a real variable and $n$ is a positive integer,then the value of $f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\ldots+\frac{f^{(n-1)}(0)}{(n-1) !}$ is

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