$a_n$ denotes the number of $n$-digit numbers made from $5,6,7,8,9$ which have an odd number of $7$'s.
Let $b_n$ denote the number of $n$-digit numbers made from $5,6,7,8,9$ which have an even number of $7$'s. Let there be $n$-spots for the $n$-digit number $\\\\...\\$, the first $(n-1)$ spots can be filled in $a_{n-1}$ ways which is followed by the last spot which can be filled in $4$ ways, if the last spot is a $7$ then we can fill the remaining $(n-1)$ spots in $b_{n-1}$ ways.
Thus, we have the recurrence relation $a_n=4a_{n-1}+b_{n-1}$. Similarly, for $b_n$ the first $(n-1)$ spots can be filled in $b_{n-1}$ ways which is followed by the last spot which can be filled in $4$ ways, if the last spot is a $7$ then we can fill the remaining $(n-1)$ spots in $a_{n-1}$ ways, thus we have the recurrence relation $b_n=4b_{n-1}+a_{n-1}$.
We can simply compute $a_1$ and $b_1$ without the recurrence. $a_1=1$, $b_1=4$.
Now, $(a_n+b_n)=5(a_{n-1}+b_{n-1})$ and $(b_n-a_n)=3(b_{n-1}-a_{n-1})$. From this we know that $(a_n+b_n)$ forms a G.P. with common ratio $5$ and first term $5$ and $(b_n-a_n)$ forms a G.P. with common ratio $3$ and first term 3.
Thus, $(a_n+b_n)=5^n$ and $(b_n-a_n)=3^n$. This gives us that $a_n=\frac{5^n-3^n}{2}$. Thus, $2a_{100}=5^{100}-3^{100}$.