Explicit formula of a sequence

Given the recurrent relation $a_n + 2a_{n-2} = 3a_{n-1}$ , we can rewrite it as:

$\begin{aligned} a_n - 3a_{n-1} + 2a_{n-2} & = 0 & \small \color{#3D99F6} \text{Its characteristic polynomial is:} \\ r^2 - 3r + 2 & = 0 \\ (r-1)(r-2) & = 0 \\ \implies a_n & = c_1(1^n) + c_2(2^n) & \small \color{#3D99F6} \text{where }c_1 \text{ and }c_2 \text{ are constants.} \\ c_1 + 2c_2 & = a_1 = 10 \quad ...(1) \\ c_1 + 4c_2 & = a_2 = 13 \quad ...(2) \\ (2)-(1): \quad c_2 & = \frac 32 \\ (1) : \quad c_1 & = 7 \\ \implies a_n & = 7 + \frac 32\cdot 2^n \\ & = 10 + 3(2^{n-1}-1) \end{aligned}$

Therefore, $A+B+C+2D = 10+3+2+2(1) = \boxed{17}$ .

Subtracting $a_1$ from $a_2$ we get $BC(C-1)=3C^D$ . So $C$ is not equal to $1$ . Also $B$ is not zero. Using the recurrance relation, we get $C^2-3C+2=0$ . Since C is not 1, therefore $C=2$ . Hence $D=1$ and $B=3$ . From the value of $a_1$ we get $A=10$ . Therefore $A+B+C+2D=17$

We are told that $\:a_{n+2}+2a_n=3a_{n+1}$ , that expression can be rearranged as follows: $a_{n+2}+2a_n=3a_{n+1} \rightarrow a_{n+2}+2a_n=a_{n+1}\:+\:2a_{n+1}\:\rightarrow \:a_{n+2}-a_{n+1}=2a_{n+1}-2_{a_n\:\rightarrow }\:a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_n\right)$ The last result, put simply, means that the difference between every two consecutive terms is twice as large as the difference between the smaller term and the one before it. So, if we were to put $n=1$ , we would get $a_3-a_2=2\left(a_2-a_1\right)$ . Since we're told $a_1=10$ & $a_2=13$ , that would lead us to $a_3-13=2\left(13-10\right)\:\rightarrow \:a_3=19$ . Continuing this rule, we could easily compute more terms in the series:

$a_4-a_3=2\left(a_3-a_2\right)\rightarrow \:a_4-19=12\:->\:a_4=31$

$a_5-a_4=2\cdot 12\:\rightarrow \:a_5-31=24\:\rightarrow \:a_5=55$

$a_6-a_5=2\cdot 24\:\rightarrow a_6-55=48\:\rightarrow \:a_6=103$

It can go on and on, but there is a pattern here:

$a_1=10$

$a_2=13=10+3$

$a_3=19=10+3+6$

$a_4=31=10+3+6+12$

$a_5=55=10+3+6+12+24$

$a_6=103=10+3+6+12+24+48$

Every time we advance in the sequence, we essentially take the value of the previous term and add to it twice the difference of the two previous terms. The difference of the first two terms is 3, so we start with 10, and then for every following term add the first $n-1$ terms of the geometric sequence which starts with 3 and has a ratio of 2. We can write it like that:

$a_n=10+\left(3+6+12+...\:+3\cdot 2^{n-1}\right)$

*I'm aware that this is far from a proper proof. If someone wants to write an organized proof using mathematical induction it would be awesome.

We can express $3+6+12+...\:+3\cdot 2^{n-1}$ using the formula for the sum of a geometric progression. The first term is 3, the ratio is 2, and we want to calculate the sum of the first $n-1$ terms, so: $3+6+12+...\:+3\cdot 2^{n-1}=\frac{3\left(2^{n-1}-1\right)}{2-1}=3\left(2^{n-1}-1\right)$ and therefore: $a_n=10+3\left(2^{n-1}-1\right)$

$A=10,\:B=3,\:C=2,\:D=1,\\:\rightarrow \:A+B+C+2D\:=\:17$