Question

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

• A. -2021
• B. 2022
• C. 2023
• D. -2023

Answer 1

The Correct Option is B

We are given that $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$. 

We want to find $f(2023)$.

Let $(1+x)(1-x)^n = \sum_{k=0}^{n+1} a_k x^k$. Then $f(n) = a_n$.

Expanding $(1+x)(1-x)^n$, we have:

$(1+x)(1-x)^n = (1-x)^n + x(1-x)^n = \sum_{k=0}^n \binom{n}{k} (-x)^k + x \sum_{k=0}^n \binom{n}{k} (-x)^k$

$= \sum_{k=0}^n \binom{n}{k} (-1)^k x^k + \sum_{k=0}^n \binom{n}{k} (-1)^k x^{k+1}$

The coefficient of $x^n$ in the expansion is:

$\binom{n}{n} (-1)^n + \binom{n}{n-1} (-1)^{n-1}$

Thus, $f(n)$ can be expressed as:

$f(n) = (-1)^n + n (-1)^{n-1} = (-1)^n + (-1)^{n-1} n$

$f(n) = (-1)^n - n (-1)^n = (-1)^n (1 - n)$

Substituting $n = 2023$, we get:

$f(2023) = (-1)^{2023} (1 - 2023) = -1 (-2022) = 2022$.

Final Answer:
The final answer is ${2022}$.