Prove that any palindrome with an even number of digits is divisible by 11 Confusing myself here, need some clarification..
First, we consider the palindrome $abccba$. We can see this can be written as
$$a(10^5 + 10^0) + b(10^4 + 10^1) + c(10^3 + 10^2) = a(10^5 + 1) + 10b(10^3 + 1) + 100c(10 + 1)$$ So essentially we see that all palindromes of even digits can be written in the form $x(10^{2k+1} + 1)$, i.e. we must show that any number of the form $(10^{2k+1} + 1)$ is divisible by $11$.
Base case: $10^{2(0)+1} + 1 = 10 + 1 = 11$, which is clearly divisible by $11$.
Induction hypothesis: Assume that $(10^{2k+1} + 1)$ is divisible by $11$, we work to show that $(10^{2(k+1)+1} + 1)$ is divisible by $11$. That is, $(10^{2k+3} + 1) = (10^2\cdot10^{2k+1} + 1) = \dots$
Where am I going wrong?

Also, notice that in your inductive proof:
$10^{2k+3} + 1$ = $10^{(2k+2)+1}+1$ = $10^{2\overline{k} +1}+1$.
