The expression 2^{4n} - 15n - 1, where n in N | vedclass

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(B) We have the expression $2^{4n} - 15n - 1$. Since $2^4 = 16$, we can write $2^{4n} = (16)^n = (1 + 15)^n$. Using the Binomial Theorem, $(1 + 15)^n

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$225$

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(B) We have the expression $2^{4n} - 15n - 1$. Since $2^4 = 16$, we can write $2^{4n} = (16)^n = (1 + 15)^n$. Using the Binomial Theorem, $(1 + 15)^n = {}^{n}C_{0} + {}^{n}C_{1}(15) + {}^{n}C_{2}(15)^2 + \dots + {}^{n}C_{n}(15)^n$. Substituting this into the expression: $2^{4n} - 15n - 1 = (1 + 15n + {}^{n}C_{2} \cdot 15^2 + \dots + {}^{n}C_{n} \cdot 15^n) - 15n - 1$. Simplifying the terms, the $1$ and $15n$ cancel out: $2^{4n} - 15n - 1 = {}^{n}C_{2} \cdot 15^2 + {}^{n}C_{3} \cdot 15^3 + \dots + {}^{n}C_{n} \cdot 15^n$. Factoring out $15^2 = 225$: $2^{4n} - 15n - 1 = 225 \cdot ({}^{n}C_{2} + {}^{n}C_{3} \cdot 15 + \dots + {}^{n}C_{n} \cdot 15^{n-2})$. Thus, the expression is divisible by $225$ for all $n \in N$ where $n \ge 2$. For $n=1$, the expression is $2^4 - 15(1) - 1 = 16 - 15 - 1 = 0$, which is divisible by $225$.

The expression $2^{4n} - 15n - 1$, where $n \in N$ (the set of natural numbers), is divisible by

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