For each n in N, 2^{3n}-1 is divisible by, wh | vedclass

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Question

(A) We know that $2^{3n} = (2^3)^n = 8^n$. Since $8^n = (1+7)^n$, we can expand this using the Binomial Theorem: $(1+7)^n = 1 + {}^{n}C_1(7) + {}^{n}C

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) We know that $2^{3n} = (2^3)^n = 8^n$. Since $8^n = (1+7)^n$, we can expand this using the Binomial Theorem: $(1+7)^n = 1 + {}^{n}C_1(7) + {}^{n}C_2(7^2) + \dots + {}^{n}C_n(7^n)$. Therefore, $2^{3n} - 1 = (1 + {}^{n}C_1(7) + {}^{n}C_2(7^2) + \dots + {}^{n}C_n(7^n)) - 1$. $2^{3n} - 1 = 7({}^{n}C_1 + {}^{n}C_2(7) + \dots + {}^{n}C_n(7^{n-1}))$. This expression is clearly divisible by $7$.

For each $n \in N$, $2^{3n}-1$ is divisible by, where $N$ is the set of natural numbers.

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