I missed one of my class today morning where in Recurrence Relation was conducted. In the solution to one problem the following stage is reached:
$C_0+\frac{C_1}3+\frac{C_2}9=0,\,C_0+\frac{C_1}4+\frac{C_2}{16}=0$, and $2(C_0+C_1+C_2)=6$.
Solving these equations, we get $C_0=\frac12$, $C_1=\frac{-7}2$, $C_2=6$.
I am unable to understand the last line.
(Original image here.)
It’s a standard problem of solving three linear equations in three unknowns; you should know at least one way to do this. Here is one extremely elementary approach. Multiply the first by $9$ and the second by $16$, and divide the third by $2$: you get
$$\left\{\begin{align*} &9C_0+3C_1+C_2=0\\ &16C_0+4C_1+C_2=0\\ &C_0+C_1+C_2=3\;. \end{align*}\right.$$
Subtract the third equation from each of the other two:
$$\left\{\begin{align*} &8C_0+2C_1=-3\\ &15C_0+3C_1=-3\;. \end{align*}\right.$$
Divide the first by $2$ and the second by $3$:
$$\left\{\begin{align*} &4C_0+C_1=\frac{-3}2\\ &5C_0+C_1=-1\;. \end{align*}\right.$$
Subtract the first from the second to get $C_0=\frac12$. Substitute that into either of the last two equations to find that $C_1=\frac{-7}2$. Then substitute these values into $C_0+C_1+C_2=3$ to find that $C_2=6$.
You have three simultaneous equations in three unknowns. You can write the last as $C_0=3-C_1-C_2$ and substitute that into the first two. That will get you down to two equations in two unknowns. Solve one of them for $C_1$, substitute into the other, and you will have one equation in $C_2$. Solve it, plug that value into the others and you will get there. It doesn't really matter that this came from a recurrence relation. There are also matrix methods for solving the three equations.
Nothing actually appears to be circled, but here's a bit of an explanation of what's going on:
You're told that the general solution to the recurrence relation $C_0 a_n + C_1 a_{n-1} + C_2 a_{n-2} = f(n)$ is $a_n = 3^n + 4^n + 2$. You're also given the information that we're interested in the specific recurrence relation where $f(n) = 6$, i.e. the relation is just $C_0 a_n + C_1 a_{n-1} + C_2 a_{n-2} = 6$. And you're being asked to find the coefficients $C_0, C_1, C_2$.
Putting the given solution into that equation gives you $C_0\left(3^n+4^n+2\right) + C_1\left(3^{n-1}+4^{n-1}+2\right) + C_2\left(3^{n-2}+4^{n-2}+2\right) = 6$. Note that this equation must be true for all $n > 2$, which means that we are looking for values of the coefficients to make it identically true (i.e. so it doesn't depend on a particular value of $n$). That will be the case when we make all the $3^n$ terms and $4^n$ terms disappear, and all the constant terms cancel out with the $6$. So, by using the fact that $3^{n-1} = \frac{3^n}{3}$ and similarly for the other terms, we have:
$\begin{eqnarray}C_0 \left(3^n + 4^n + 2 \right) + C_1 \left( \frac{3^n}{3} + \frac{4^n}{4} + 2 \right) + C_2 \left( \frac{3^n}{9} + \frac{4^n}{16} + 2 \right) & = & 6\\ 3^n \left( C_0 + \frac{C_1}{3} + \frac{C_2}{9} \right) + 4^n \left( C_0 + \frac{C_1}{4} + \frac{C_2}{16} \right) + \left( C_0 + C_1 + C_2 \right) & = & 6\end{eqnarray}$
The right hand side has nothing that depends on $3^n$, so the term on the left that depends on $3^n$ must disappear, i.e. $C_0 + C_1 / 3 + C_2 / 9 = 0$, and similarly for the $4^n$ term. Then we also have that the constant term on the right is $6$, so the bit on the left that doesn't depend on $n$ must also be $6$, so $C_0 + C_1 + C_2 = 6$, and then we just have three linear equations in 3 unknowns to solve.
