For all integers n geq 1, which of the follow | vedclass

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(B) Using the binomial expansion, we have $4^n = (1+3)^n$. By the binomial theorem, $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots + x^n$. Substitut

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$4^n-3n-1$

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(B) Using the binomial expansion, we have $4^n = (1+3)^n$. By the binomial theorem, $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots + x^n$. Substituting $x=3$, we get $4^n = 1 + 3n + \frac{n(n-1)}{2!} \cdot 3^2 + \dots + 3^n$. Rearranging the terms, we get $4^n - 3n - 1 = \frac{n(n-1)}{2} \cdot 9 + \dots + 3^n$. Since every term on the right side contains a factor of $9$ (or a higher power of $3$ which is divisible by $9$), the expression $4^n - 3n - 1$ is divisible by $9$ for all integers $n \geq 2$. For $n=1$, $4^1 - 3(1) - 1 = 0$, which is divisible by $9$ (as $0 = 9 	imes 0$). Thus, $4^n - 3n - 1$ is divisible by $9$ for all $n \geq 1$.

For all integers $n \geq 1$, which of the following is divisible by $9$?

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