If f(n) is the coefficient of x^n in the expa | vedclass

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(B) The given expression is $(1+x)(1-x)^n = (1-x)^n + x(1-x)^n$. $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$. $f(n) = (	ext

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$2020$

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(B) The given expression is $(1+x)(1-x)^n = (1-x)^n + x(1-x)^n$. $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$. $f(n) = (	ext{coefficient of } x^n 	ext{ in } (1-x)^n) + (	ext{coefficient of } x^{n-1} 	ext{ in } (1-x)^n)$. Using the binomial expansion $(1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k$. The coefficient of $x^n$ in $(1-x)^n$ is $\binom{n}{n}(-1)^n = (-1)^n$. The coefficient of $x^{n-1}$ in $(1-x)^n$ is $\binom{n}{n-1}(-1)^{n-1} = n(-1)^{n-1}$. Thus,$f(n) = (-1)^n + n(-1)^{n-1} = (-1)^n - n(-1)^n = (-1)^n(1-n)$. For $n = 2021$,$f(2021) = (-1)^{2021}(1-2021) = (-1)(-2020) = 2020$.

If $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$,then $f(2021)=$

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