Closed form for $T_n$ if $T_{n+3}=2 T_{n+2}+2 T_{n+1}-T_{n}$ Consider the recurrence relation $$T_{n+3}=2 T_{n+2}+2 T_{n+1}-T_{n}$$ with first three terms as:
$T_{1}=20, T_{2}=12, T_{3}=70$
Find the closed form expression for $T_n$.
My try:
Since it is a constant coefficient difference equation, the auxiliary equation is:
$$\lambda^3-2\lambda^2-2\lambda+1=0$$ whose roots are:
$-1,\frac{3\pm\sqrt{5}}{2}$
Thus we have:
$$T_{n}=A(-1)^{n}+B\left(\frac{3+\sqrt{5}}{2}\right)^{n}+C\left(\frac{3-\sqrt{5}}{2}\right)^{n}$$
where the constants $A,B,C$ to be determined by values of $T_1,T_2,T_3$ but its becoming too tedious to solve. Any alternate approach?


$$T_{n}=A(-1)^{n}+B\left(\frac{3+\sqrt{5}}{2}\right)^{n}+C\left(\frac{3-\sqrt{5}}{2}\right)^{n}$$
Where the constants $A,B,C$ to be determined by values of $T_1,T_2,T_3$ but its becoming too tedious to solve. Any alternate approach?

The approach is correct, and can be made less tedious by noting that:

*

*It is easier to  shift the powers $\,T_n=A'(-1)^{n-1}+B'\left(\frac{3+\sqrt{5}}{2}\right)^{n-1}+C'\left(\frac{3-\sqrt{5}}{2}\right)^{n-1}\,$ since the sequence starts with index $\,1\,$.


*All terms in the sequence are integers, so $\,B',C'\,$ must be rational conjugates, therefore $\,A'=a\,$ and $\,B',C'=b \pm c\sqrt{5}\,$ where $\,a,b,c \in \mathbb Q\,$.
Then:

*

*$\require{cancel} \color{red}{20} = T_1 = a+b+\cancel{c\sqrt{5}}+b-\cancel{c\sqrt{5}} \color{red}{= a + 2 b}$


*$\color{red}{24} = 2 \cdot T_2 = -2a + (b+c\sqrt{5})(3+\sqrt{5}) + (b-c\sqrt{5})(3-\sqrt{5})$ $= -2a+3b+5c+\cancel{(b+3c)\sqrt{5}}+3b+5c-\cancel{(b+3c)\sqrt{5}}$ $\color{red}{= -2a + 6b + 10c}$


*$\color{red}{280} = 4 \cdot T_3 = 4a + (b+c\sqrt{5})(3+\sqrt{5})^2 + (b-c\sqrt{5})(3-\sqrt{5})^2$ $= 4a + (b+c\sqrt{5})(14 + 6 \sqrt{5}) + (b-c\sqrt{5})(14-6\sqrt{5})$ $= 4a+14b+30c+\cancel{(6b+14c)\sqrt{5}}+14b+30c-\cancel{(6b+14c)\sqrt{5}}$ $\color{red}{= 4a + 28 b + 60 c}$
Solving the system for $\,a,b,c\,$ gives $\,a = \dfrac{54}{5}, b = \dfrac{23}{5}, c = \dfrac{9}{5}\,$.
