<s>[INST] Sum of $a_0+a_1+\cdots + a_n$ If $a_0=2 , a_1=5 $and for $n>1 $
$$a_n=5a_{n-1}-6a_{n-2}$$
then
$$a_0+a_1+\cdots+a_n=?$$
I know telescoping series but 
$$a_n=5(a_{n-1}-a_{n-2})-a_{n-2}$$
Thanks for your help.
Mani
 [/INST] Clearly
$$ a_n-2a_{n-1}=3(a_{n-1}-2a_{n-2}). $$
So
$$ a_n-2a_{n-1}=3^{n-1}(a_{1}-2a_{0})=3^{n-1}. \tag{1}$$
Similarly
$$ a_n-3a_{n-1}=2^{n-1}(a_{1}-2a_{0})=-2^{n-1}. \tag{2}$$
(1)(2) give
$$ a_n=3^n+2^n.$$
So
$$ a_0+a_1+a_2+\cdots+a_n=(1+3+3^2+\cdots+3^n)+(1+2+2^2+\cdots+3^n)=\frac12\cdot(3^{n+1}-1)+(2^{n+1}-1).$$
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