Came across this problem recently, to which my textbook gave easily the most disappointing solution you could think of.
Consider the following recurrence, $$u_{n+2}=5u_{n+1}-6u_{n}$$ $a)$ show that for any sequence, $u_n=3^np+2^nq$, this recurrence is satisfied, $b)$ If $u_1=5$ and $u_2=12$, then express $u_n$ in terms of $n$ only, and $c)$ determine the number of digits in $u_{100}$
So, for a if we consider that $$u_n=3^np+2^nq~~~\implies-6u_{n}=-6[3^np]-6[2^nq]$$then$$u_{n+1}=3[3^np]+2[2^nq]~~~\implies5u_{n+1}=15[3^np]+10[2^nq]$$and$$u_{n+2}=9[3^np]+4[2^nq]$$ $$u_{n+2}=15[3^np]-6[3^np]+10[2^nq]-6[2^nq]$$ $$u_{u+2}=\left[15[3^np]+10[2^nq]\right]+\left[-6[3^np]-6[2^nq]\right]$$ $$u_{n+2}=5u_{n+1}-6u_n$$
Now, for b, we know that when $n=1$, $$3p+2q=5~~~\implies9p+6q=15\tag{1}$$and that, when $n=2$, we have $$9p+4q=12\tag{2}$$now, subtracting $(2)$ from $(1)$ yields $$2q=3~~~\implies q=\frac{3}{2}~~~\implies p=\frac{2}{3}$$which ultimately leaves us with$$u_n=2[3^{n-1}]+3[2^{n-1}]$$now this last part is where I got let down. I tried taking a rigorous approach, looking for a pattern of digits, and found that $u_1$ has $1$ digit, $u_{2,~3,~4}$ has $2$ digits, and after this, each pair of $u_{n}$ has $1$ digit more than the prior pair. Using this idea, I removed the initial $4$ $u_n$'s that were not in a pair, i.e. everything up to $u_4$, and found that there are $(100-4)/2=48$ pairs after $u_4$, and $u_{100}$ lands on the last value of one of these pairs. Given that the first pair of $u_n$ after $u_4$ has $3$ digits, I followed on this idea and concluded that the $48^{th}$ pair would have $51$ digits, and concluded that $u_{100}$ has $51$ digits. As you may have guessed this is incorrect, however, my textbook simply gave the solution as one to an equation that's just been plugged into a calculator, and whose exponent when in standard form was analysed. I am however, wondering if there is a more rigorous way to approach this problem, other than just plugging values into a calculator.
Also I'm not sure whether the original tags are correct, feel free to adjust them accordingly.
Any responses are appreciated. Thank you.
Since $2^{99}\ll3^{99}$, it's $$\left[\log_{10}(2\cdot3^{99}+3\cdot2^{99})\right]+1=\left[\log2+99\log3\right]+1=48$$