For each n∈N,23n−1is divisible by - Infinity Learn by Sri Chaitanya

A
8
B
16
C
32
D
none of these

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 divides $2^{3 n} - 1$ for all $n \in N$ .

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