$$a_{n+2}-6a_{n+1}+9a_n=3(2^n)+7{n \choose 3} , a_2=4,a_1=1,n\ge3$$
my try:
(1) $a_n=a^{h}_n+a^{p}_n$
(2) solving homogeous part: $x^2-6x+9=0$ so $x=3$. then we have $a^{h}_n=(k_1+nk_2)3^n$.
(3) solving particular part: suppose $a_n=A2^n$ then $4A-12A+9A=3+7\frac{{n \choose 3}}{2^n}$ so $a^{p}_n=3(2^n)+7{n \choose 3}$.
(4) finding $k_1$ and $k_2$: $a_n=(k_1+nk_2)3^n+3(2^n)+7{n \choose 3}$
$a_1=1=(k_1+k_2)3+6+7{1 \choose 3}$ and $a_2=4=(k_1+2k_2)9+12+7{2 \choose 3}$ but I don't know ${1 \choose 3}$ and ${2 \choose 3}$.
what should I do?
