Solution to 25. Solve recurrence relation an+3=3an+2+4an+1-12an for n20 with a0-0,al--11,a2--15 - Sikademy
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25. Solve recurrence relation an+3=3an+2+4an+1-12an for n20 with a0-0,al--11,a2--15

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Let us solve recurrence relation a_{n+3}=3a_{n+2}+4a_{n+1}-12a_n with a_0=0,a_1=11,a_2=15. Firstly, let us solve its characteristic equation k^3=3k^2+4k-12, which is equivalent to (k-2)(k^2-k-6)=0, and hence to (k-2)(k+2)(k-3)=0. We conclude that its roots are k_1=2,\ k_2=-2,\ k_3=3. The general solution is a_n=c_12^n+c_2(-2)^n+c_33^n. It follows that 0=a_0=c_1+c_2+c_3,\ 11=a_1=2c_1-2c_2+3c_3,\ 15=a_2=4c_1+4c_2+9c_3. The system \begin{cases}c_1+c_2+c_3=0\\ 2c_1-2c_2+3c_3=11\\ 4c_1+4c_2+9c_3=15\end{cases} is equivalent to the system \begin{cases}c_1+c_2+c_3=0\\ -4c_2+c_3=11\\ 5c_3=15\end{cases}, and hence


\begin{cases}c_1=-1\\ c_2=-2\\ c_3=3\end{cases}.


Therefore, the solution is a_n=-2^n-2(-2)^n+3\cdot3^n=-2^n+(-2)^{n+1}+3^{n+1}.


If n=20, then a_{20}=-2^{20}+(-2)^{21}+3^{21}=-2^{20}-2\cdot 2^{20}+3^{21}=3^{21}-3\cdot 2^{20}.

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