Using the principle of mathematical induction, prove each of the following

To Prove:

$x^{2 n-1}-1$ is divisible by $x-1$

Let us prove this question by principle of mathematical induction (PMI) for all natural numbers

Let $\mathrm{P}(\mathrm{n}):^{2 n-1}-1$ is divisible by $x-1$

For n = 1

$\mathrm{P}(\mathrm{n})$ is true since $x^{2 n-1}-1=x^{2-1}-1=(x-1)$

Which is divisible by x - 1

Assume P(k) is true for some positive integer k , ie,

$=x^{2 k-1}-1$ is divisible by $x-1$

Let $x^{2 k-1}-1=m \times(x-1)$, where $m \in \mathbf{N} \ldots(1)$

We will now prove that P(k + 1) is true whenever P( k ) is true

Consider

$=x^{2(k+1)-1}-1$

$=x^{2 k-1} \times x^{2}-1$

$=x^{2}\left(x^{2 k-1}\right)-1$

$=x^{2}\left(x^{2 k-1}-1+1\right)-1$ [Adding and subtracting 1]

$=x^{2}(\mathrm{~m} \times(\mathrm{x}-1)+1)-1$ [Using 1]

$=x^{2}(\mathrm{~m} \times(\mathrm{x}-1))+x^{2} \times 1-1$

$=x^{2}(\mathrm{~m} \times(\mathrm{x}-1))+x^{2}-1$

$=x^{2}(\mathrm{~m} \times(\mathrm{x}-1))+\left(x^{1}-1\right)(\mathrm{x}+1)$

$=(x-1)\left\{m x^{2}+(x+1)\right\}$, which is factor of $(\mathrm{x}-1)$

Therefore, P (k + 1) is true whenever P(k) is true

By the principle of mathematical induction, P(n) is true for all natural numbers, ie, N.

Hence proved.