If n is a positive integer and f(n) is the co | vedclass

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(B) The expression is $(1+x)(1-x)^n$. The coefficient of $x^n$ in $(1+x)(1-x)^n$ is the sum of the coefficient of $x^n$ in $(1-x)^n$ and the coefficie

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$2022$

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(B) The expression is $(1+x)(1-x)^n$. The coefficient of $x^n$ in $(1+x)(1-x)^n$ is the sum of the coefficient of $x^n$ in $(1-x)^n$ and the coefficient of $x^{n-1}$ in $(1-x)^n$. Using the binomial expansion $(1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$,the coefficient of $x^k$ is $\binom{n}{k}(-1)^k$. Thus,$f(n) = \binom{n}{n}(-1)^n + \binom{n}{n-1}(-1)^{n-1}$. $f(n) = 1 \cdot (-1)^n + n \cdot (-1)^{n-1}$. For $n = 2023$: $f(2023) = (-1)^{2023} + 2023 \cdot (-1)^{2022}$. $f(2023) = -1 + 2023(1) = 2022$.

If $n$ is a positive integer and $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$,then $f(2023) = $

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