(2^{3n} - 1) will be divisible by (forall n i | vedclass

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(C) We have $2^{3n} = (2^3)^n = 8^n$. Since $8 = (1 + 7)$, we can write $8^n = (1 + 7)^n$. Using the Binomial Theorem, $(1 + 7)^n = {^nC_0} + {^nC_1}(

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$7$

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(C) We have $2^{3n} = (2^3)^n = 8^n$. Since $8 = (1 + 7)$, we can write $8^n = (1 + 7)^n$. Using the Binomial Theorem, $(1 + 7)^n = {^nC_0} + {^nC_1}(7) + {^nC_2}(7^2) + \dots + {^nC_n}(7^n)$. Since ${^nC_0} = 1$, we have $8^n = 1 + 7({^nC_1} + {^nC_2}(7) + \dots + {^nC_n}(7^{n-1}))$. Subtracting $1$ from both sides, $8^n - 1 = 7({^nC_1} + {^nC_2}(7) + \dots + {^nC_n}(7^{n-1}))$. This expression is clearly divisible by $7$ for all $n \in N$.

$(2^{3n} - 1)$ will be divisible by $(\forall n \in N)$

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