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

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

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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,$4^n = 1 + n(3) + \frac{n(n-1)}{2!} 3^2 + \frac{n(n-1)(n-2)}{3!} 3^3 + \dots$. This simplifies to $4^n = 1 + 3n + 9 \left[ \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!} (3) + \dots \right]$. Rearranging the terms,we get $4^n - 3n - 1 = 9 \left[ \frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!} (3) + \dots \right]$. Since the expression inside the bracket is an integer for all $n \geq 1$,the expression $4^n - 3n - 1$ is divisible by $9$.

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

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