For each n∈N,$$23n$$−$$1$$ is divisible by

We have $$23n$$−$$1=23n$$−$$1=(1+7)n$$−$$1=1+nC1(7)+nC272+$$…$$+nCn7n$$−$$1=7mWhere$$ $$m=nC1+nC2(7)+$$…$$+nCn7n$$−$$1is$$ a positive integer.Thus, $$7$$ divided $$23n$$−$$1$$ for all n∈N

Binomial theorem for positive integral Index

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Question

For each $n \in N, 2^{3 n} - 1$ is divisible by

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Solution

We have
$\begin{array}{l} 2^{3 n} - 1 = {(2^{3})}^{n} - 1 = (1 + 7)^{n} - 1 \\ = 1 +^{n} C_{1} (7) +^{n} C_{2} (7^{2}) + \dots +^{n} C_{n} (7^{n}) - 1 \\ = 7 m \end{array}$
Where $m =^{n} C_{1} +^{n} C_{2} (7) + \dots +^{n} C_{n} (7^{n - 1})$
is a positive integer.
Thus, $7$ divided $2^{3 n} - 1$ for all $n \in N$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

For each n∈N,23n−1 is divisible by

We have 23n−1=23n−1=(1+7)n−1=1+nC1(7)+nC272+…+nCn7n−1=7mWhere m=nC1+nC2(7)+…+nCn7n−1is a positive integer.Thus, 7 divided 23n−1 for all n∈N

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For each $n \in N, 2^{3 n} - 1$ is divisible by