Prove that any palindrome with an even number of digits is divisible by 11
Hint: Notice that $10^k=(-1)^k \pmod{11}$. So, for $k$ odd and $k'$ even, $10^{k} + 10^{k'}=0\pmod{11}$.
Also, notice that in your inductive proof:
$10^{2k+3} + 1$ = $10^{(2k+2)+1}+1$ = $10^{2\overline{k} +1}+1$.