Answer
Let $P(n)$ denote the proposition: "$21$ divides $4^{n+1} + 5^{2k +1}$ whenever $n$ is a positive integer."
$P(1)$ is true, since $4^{(1)+1} +5^{2(1)-1} = 21$.
Assume that $P(k)$ is true for an arbitrary integer $k$. Then
\begin{align*}
4^{(k+1)+1} + 5^{2(k+1)-1} &= 4^{k+2} + 5^{2k+1} \\
&= 4(4^{k+1}) +25(5^{2k-1}) \\
&= 25(4^{k+1} + 5^{2k-1}) - 21(4^{k+1})
\end{align*}
Each of the bracketed terms is divisible by $21$, so $P(k+1)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - $21$ divides $4^{n+1} + 5^{2k +1}$ whenever $n$ is a positive integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "$21$ divides $4^{n+1} + 5^{2k +1}$ whenever $n$ is a positive integer."
Next, we must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $4^{n+1} + 5^{2n-1} = 4^{(1)+1} +5^{2(1)-1} = 16 + 5 =21$. $21$ is divisible by $21$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step, Assume that $P(k)$ is true; that is, we assume that $21$ divides $4^{k+1} + 5^{2k-1}$ for an arbitrary positive integer $k$. Now we must show that this implies that $P(k+1)$ is also true.
\begin{align*}
4^{(k+1)+1} + 5^{2(k+1)-1} &= 4^{k+2} + 5^{2k+1} \\
&= 4(4^{k+1}) +25(5^{2k-1}) \\
&= 25(4^{k+1}) - 21(4^{k+1}) + 25(5^{2k-1}) \\
&= 25(4^{k+1} + 5^{2k-1}) - 21(4^{k+1})
\end{align*}
The first bracketed term is divisible by $21$, via our inductive hypothesis; the second bracketed term is trivially divisible by $21$. This means that the entire expression must also be divisible by $21$, and that $P(k+1)$ is true. Therefore by method of mathematical induction $P(n)$ is true - $21$ divides $4^{n+1} + 5^{2k +1}$ whenever $n$ is a positive integer.