proof writing - Prove by induction that $3^{4n + 2} + 1$ is divisible by $10$

Prove by induction: $3^{(4n+2)} + 1$ is divisible by $10$.

My basic step: $3^{(4n+2)} + 1$, where $n = 1$ gives me $3^6 + 1 = 730$, which is divisible by $10$. However, then I have to do the induction hypothesis and I am kind of stuck because I do not have an equality. How do I finish proving this by induction?

Many thanks.

Edit: I am thinking of creating a formula which involves $10n$? Would this be correct?

Answer

$f(n): 3^{4n+2}+1$

STEP-$1$:

$f(1): 3^{4+2}+1 = 730$, which is divisible by $10$. Hence $f(1)$ holds true.

STEP-$2$:

Now let $n=k$, i.e., $f(n)=f(k)$ hold true .Hence, $f(k) = 3^{4k+2}+1$ is divisible by $10$.

Now we just need to prove that the criteria is satisfied for $n=k+1$.

STEP-$3$:

$$f(k+1) = 3^{4(k+1)+2}+1$$ $$ = 3^{4k+2}.3^{4}+1$$ $$ = 3^{4k+2}.(80+1)+1$$ $$ = (3^{4k+2}.80+3^{4k+2}.1)+1$$ $$ = 3^{4k+2}.80+(3^{4k+2}.1+1)$$

The first term is clearly divisible by 10. The second and third term are together divisible by 10 (from our assumption in step-2). So $f(n)=f(k+1)$ holds true.

Hence by induction $3^{4n+2}+1$ be divisible by $10$.