First slide

Binomial theorem for positive integral Index

Question

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

Moderate

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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