I understand quite well the steps for proof by induction. It is just that I'm doing it a little differently to the text book and want to confirm if it is valid.
The question is prove by induction that $3^{2n}+11$ is divisible by $4$
Let $f(n) = 3^{2n}+11$
The method shown in the text book is to take $f(k+1)-f(k)$ to eventually get $f(k+1) = f(k) + 4(2(3^{2k}))$, which is of course divisible by $4$
But, it seems algebraically easier if I was to take $f(k+1)-9f(k)$ to give $f(k+1) = 9f(k)-88$. Since $f(k)$ and $88$ are both divisible by $4$, $f(k+1)$ is divisible by $4$.
Is this second method still valid? Do I have to take $f(k+1)-f(k)$ or can I take away any multiple of $f(k)$ as long as I can show that $f(k+1)$ is divisible by $4$?
Thanks for your help.
Nice idea, and your method is valid because:
If $\,4$ divides $\,f(k+1)-9f(k)\,$ then $\,f(k+1)\,$ and $\,9f(k)\,$ can be either both divisible or both not divisible by $4$.
Now consider the base case where $\,k=1$. We have $\,f(1)=20$, which is divisible by $4$, and so is $\,9f(1)$. Thus, $\,f(2)\,$ is also divisible by $4$ and the base case holds.
Now, the induction has worked for all cases where $\,n\geq2$, along with the fact that $4$ divides $\,f(1)$, you can finish the proof.