Question

Using the principle of
mathematical induction show that 10^(2n-1 ) + 1 is divisible by 11 for all z

Answer 1

When $n=1$,

$10^{2\cdot1 - 1} + 1 = 10^1 + 1 = 11$

which is of course divisible by 11.

Assume this holds for $n=k$, that

$11 \mid 10^{2k - 1} + 1$

In other words,

$10^{2k - 1} + 1 = 11\ell$

for some integer $\ell$.

Use this to show the claim is true for $n=k+1$.

$10^{2(k+1) - 1} + 1 = 10^{2k + 1} + 1 \\\\ ~~~~~~~~~~~~~~~~~~~~ = 10^{2k+1} + \left(10^{2k-1} + 10^{2k-1}\right) + 1 \\\\ ~~~~~~~~~~~~~~~~~~~~ = \left(10^{2k+1} - 10^{2k-1}\right) + \left(10^{2k-1} + 1\right) \\\\ ~~~~~~~~~~~~~~~~~~~~ = 10^{2k-1} \left(10^2 - 1\right) + 11\ell \\\\ ~~~~~~~~~~~~~~~~~~~~ = 99\times10^{2k-1} + 11\ell \\\\ ~~~~~~~~~~~~~~~~~~~~ = 11\left(9\times10^{2k-1} + \ell\right)$

which is indeed divisible by 11. QED

On the off-chance you meant $10^{2^n-1}+1$, notice that $2n-1$ is odd for any integer $n$. Similarly $2^n-1$ is odd for all $n$, so the above proof actually proves this automatically.