Q.Prove the following by using the principle of mathematical induction for all $n \in N$: $10^{2n - 1} + 1$ is divisible by 11.

Let $P(n)$ be the statement: $10^{2n-1}+1$ is divisible by $11$.

Base case: For $n=1$,

divisible by $11$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$, i.e. there is an integer $m$ with

We must show $10^{2(k+1)-1}+1=10^{2k+1}+1$ is divisible by $11$.

Write $10^{2k+1}=10^2\cdot10^{2k-1}=100\cdot10^{2k-1}$:

Substitute the induction hypothesis $10^{2k-1}+1=11m$: …

Divisibility induction. Express $10^{2k+1}$ as $100\times10^{2k-1}$, write $100\big(10^{2k-1}+1\big)-99$ so the induction hypothesis substitutes in …

• Errors relating $10^{2(k+1)-1}$ to $10^{2k-1}$: the exponent jumps by $2$ (from $2k-1$ to $2k+1$), i.e. multiplication by $10^2=100$, not by $10$.
• Forgetting the $-100+1=-99$ correction term when rewriting $100\cdot10^{2k-1}+1$ in terms of $10^{2k-1}+1$. …