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*10^(2k-1) + 11\ell \\\\ ~~~~~~~~~~~~~~~~~~~~ = 11\left(9*10^(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.