Induction: Prove that $4^{n+1}+5^{2 n - 1}$ is divisible by 21 for all $n \geq 1$.

HINT: Let $f(n)=4^{n+1}+5^{2n-1}$; you want to show that $f(n)$ is divisible by $21$ for all $n\ge 1$.

Presumably the base case, proving that $f(1)$ is divisible by $21$, causes no trouble. For your induction step you want to show that if $f(n)$ is divisible by $21$ for some $n\ge 1$, then $f(n+1)$ is divisible by $21$. Note that this will be true if and only if $f(n+1)-f(n)$ is divisible by $21$, so you ought to look at

$$\begin{align*} f(n+1)-f(n)&=\left(4^{n+2}+5^{2n+1}\right)-\left(4^{n+1}-5^{2n-1}\right)\\ &=\left(4^{n+2}-4^{n+1}\right)+\left(5^{2n+1}-5^{2n-1}\right)\\ &=4^{n+1}(4-1)+5^{2n-1}\left(25^2-1\right)\;.\tag{1} \end{align*}$$

Now see if you can rearrange $(1)$ to get something that is demonstrably a multiple of $21$.

Ostensibly, it can be easily established using congruence.

But as induction is required,it can be done as follows:

If $f(m)=4^{m+1}+5^{2m-1}$

$f(m+1)-f(m)$

$=4^{m+2}+5^{2m+1}-(4^{m+1}+5^{2m-1})$

$=4^{m+1}(4-1)+5^{2m-1}(5^2-1)$

$=3(4^{m+1}+5^{2m-1})+21\cdot 5^{2m-1}$

$f(m+1)=21\cdot 5^{2m-1}+4\cdot f(m)$

Now, $f(1)=4^2+5^1=21,f(2)=4^3+5^3=189$

So using induction,we can show that $21\mid f(n)$ if $n\ge 1$

Alternatively, let $f(m)=a^{2m-1}+b^{m+1}$

So, $f(m+1)-f(m)=a^{2m+1}+b^{m+2}-(a^{2m-1}+b^{m+1})$ $=(b-1)(a^{2m-1}+b^{m+1})+(a^2-b)a^{2m-1}$

$\implies f(m+1)=b(a^{2m-1}+b^{m+1})+(a^2-b)a^{2m-1}=f(m)+(a^2-b)a^{2m-1}$

No,w $f(1)=(a+b^2)$

So if $(a^2-b)\mid(a+b^2), (a^2-b)\mid f(n)$ if $n\ge 1$

Here $a=5,b=4\implies a^2-b=25-4=21$ and $a+b^2=5+4^2=21$

$5^{2n-1} = 5 \cdot 25^{n-1}$. Hence, $$5^{2n-1} \equiv 5 \cdot (21+4)^{n-1} \pmod{21} \equiv 5 \cdot 4^{n-1} \pmod{21}$$

Hence, $$4^{n+1} + 5^{2n-1} \equiv 4^{n+1} + 5 \cdot 4^{n-1} \pmod{21} \equiv 21 \cdot 4^{n-1} \pmod{21} \equiv 0 \pmod{21}$$

For a proof by induction, note that if $f(n) = 4^{n+1} + 5^{2n-1}$, then $$f(n+1) - 4f(n) = 4^{n+2} + 5^{2n+1} - 4^{n+2} - 4 \cdot 5^{2n-1} = 21 \cdot 5^{2n-1}$$

Hence, $$f(n+1) = 4f(n) + 21 \cdot 5^{2n-1}$$

Hint $ $ Times $5$ it's $\rm\, 25^n\!+20\cdot 4^n\equiv\, 4^n\!- 4^n\equiv\, 0 \!\pmod{21}\ $ (easily proved by induction if need be)

Remark $\ $ More generally $\rm\: a^3\equiv -1\:\Rightarrow\: (a^2)^{n+1}\!+a^{2n-1}\! =\, a^{2n-1}(a^3+1)\equiv 0$