I tried this in a way like my assumption is that $${T_n}$$ be a string with no consecutive zeros and one's,and now calculating for starting bits in the sequence is as follows
01=$$T_{n-2}$$
02=$$T_{n-2}$$
2=$$T_{n-1}$$
so I got the answer as $${T_n}=2*T_{n-2}+T_{n-2}$$
But the given answer is $${T_n}=2*T_{n-1}+T_{n-2}$$.
There might be easier ways to think about it (combinatorics is like that), but here is my idea.
Let's think about the $T_n$ sequences of length $n$. Clearly, $T_{n-1}$ of them end in 2. Of the rest, because of the $0-1$ symmetry in the problem, half of them end in $0$ and half in $1$, which is $\frac12(T_n-T_{n-1})$.
So how many sequences of length $n+1$ will there be? Of the sequences that end in $2$, there are three allowable ways to add a trit to the end. (A trit is a ternary digit.) Of the sequences that end in $0$ or $1$, there are two allowable trits that can continue the sequence since you cannot repeat that last trit. Therefore,
$$T_{n+1}=3T_{n-1}+2\left(\frac12(T_n-T_{n-1})\right)=3T_{n-1}+T_n-T_{n-1}=T_n+2T_{n-1}$$