Question

By the method of induction prove that $\left( {{2}^{3n}}-1 \right)$ is divisible by 7.

Answer 1

To solve this question, we will assume P(n) be a statement such that ${{2}^{3n}}-1$ is divisible by 7. After proving P(1) = true by taking n = 1, we will assume that P(k) is true. Finally, from this, P(k) is a true statement and we will try to get that P(k + 1) is also true.

Complete step-by-step answer:
We have to prove that ${{2}^{3n}}-1$ is divisible by 7 with the help of mathematical induction. Let P(n) be a statement such that $P\left( n \right)={{2}^{3n}}-1$ is divisible by 7. To show n = 1 is true, P(n) = P(1) and substituting n = 1 in ${{2}^{3n}}-1.$
$\Rightarrow {{2}^{3\times 1}}-1$
$\Rightarrow {{2}^{3}}-1$
$\Rightarrow 8-1$
$\Rightarrow 7$
And as 7 is divisible by 7. Therefore, ${{2}^{3}}-1$ is divisible by 7. Hence P(1) is true as it is stated ${{2}^{3}}-1$ is divisible by 7…..(i)
Let us assume P(k) is true where n = k and P(k) is true.
$\Rightarrow {{2}^{3k}}-1$ is divisible by 7.
$\Rightarrow P\left( k \right)={{2}^{3k}}-1=7d$
It means it can be written as a multiple of 7 where d is any natural number.
$\Rightarrow {{2}^{3k}}-1=7d$
$\Rightarrow {{2}^{3k}}=7d+1.....\left( iii \right)$
where d = 1, 2, 3, 4…..
Now, we finally check for n = R + 1.
$P\left( k+1 \right)={{2}^{3\left( k+1 \right)}}-1$
For P(k + 1) to be true, we should have,
${{2}^{3\left( k+1 \right)}}-1=7d$
That is we should have, ${{2}^{3\left( k+1 \right)}}=7d+1.$
Now, ${{2}^{3\left( k+1 \right)}}={{2}^{3k}}{{.2}^{3}}$
As ${{a}^{xy}}={{a}^{x}}{{a}^{y}}$ where ‘a’ is any real number and x and y are real numbers or integers.
$\Rightarrow P\left( k+1 \right)={{2}^{3k}}{{.2}^{3}}-1$
Now from equation (iii), we had that,
${{2}^{3k}}=7d+1$
Substituting this value of ${{2}^{3k}}$ in the above equation, we have,
$P\left( k+1 \right)=\left( 7d+1 \right){{.2}^{3}}-1$
$\Rightarrow P\left( k+1 \right)=\left( 7d+1 \right)8-1$
$\Rightarrow P\left( k+1 \right)=56d+8-1$
$\Rightarrow P\left( k+1 \right)=56d+7$
$\Rightarrow P\left( k+1 \right)=7\left( 8d+1 \right)$
Hence, P(k + 1) is written as a product of 7 and (8d + 1) where d = 1, 2, 3…. Hence, we have, $P\left( k+1 \right)={{2}^{3\left( k+1 \right)}}-1$ is divisible by 7. So, P(k + 1) is true…..(iv)
So, finally from equation (i), (ii) and (iv), we have that P(1), P(k) and P(k + 1); k > 1 is true.
Hence, by mathematical induction, we have shown that ${{2}^{3n}}-1$ is divisible by 7.
Hence proved.

Note: The point where we have written that for P(k + 1) to be true, we should get ${{2}^{3\left( k+1 \right)}}-1=7d$ type. Here, this ‘d’ can also be replaced by any other natural number ‘s’ because anyways we are going to get any number with a multiple of 7. We have used ‘d’ here as it was used earlier with P(k). Anyways, it can be shifted if required. Because clearly, ${{2}^{3k}}-1\ne {{2}^{3\left( k+1 \right)}}-1.$ So, we can change the variable too.