Related Recurrence Relations

Rearranging

$\begin{cases} b_n=c_{n-1}-c_{n-2}-3a_{n-2}+2^{n-2} \phantom{cciiccc} (1)\\ a_{n-1}=2b_{n-1}+c_{n-2}+5a_{n-2}-2^{n-3} \phantom{ccc} (2)\\ c_n=a_n+6a_{n-2}-5a_{n-1}-2^{n-2} \phantom{ccccccc} (3) \end{cases}$

We can increase/decrease every indices/exponent that depends on $n$ by the same amount without changing the recurrence. Increasing by $1$ in $(2)$

$\begin{cases} b_n=c_{n-1}-c_{n-2}-3a_{n-2}+2^{n-2} \phantom{ccic} (1)\\ a_{n}=2b_{n}+c_{n-1}+5a_{n-1}-2^{n-2} \phantom{cciic} (2)\\ c_n=a_n+6a_{n-2}-5a_{n-1}-2^{n-2} \phantom{cicic} (3) \end{cases}$

Now, with $2(1)+(2)+(3)$ we get

$2b_n+a_n+c_n=2(c_{n-1}-c_{n-2}-3a_{n-2}+2^{n-2})+2b_{n}+c_{n-1}+5a_{n-1}-2^{n-2}+a_n+6a_{n-2}-5a_{n-1}-2^{n-2}$

Simplifying like terms

$c_n=3c_{n-1}-2c_{n-2}$

Using the auxiliary polynomial to find the closed form

$p(t)=t^2-3t+2$

Its roots are $t=1$ and $t=2$ , so $c_n$ is of the form

$c_n=\lambda_1+\lambda_22^n$

Using the given values of $c_0$ and $c_1$

$\begin{cases} c_0=\lambda_1+\lambda_2=\dfrac{7}{4}\\ c_1=\lambda_1+2\lambda_2=\dfrac{3}{2} \end{cases} \Longrightarrow \lambda_1=2,\lambda_2=-\dfrac{1}{4}$

$\Longrightarrow c_n=2-2^{n-2}$

Substituting back in $(3)$ and simplifying like terms

$a_n=5a_{n-1}-6a_{n-2}+2$

To find the closed form of $a_n$ I'll use a generating function

$\begin{aligned} A(x)&=\sum_{k=0}^{\infty} a_kx^k\\ &=a_0+a_1x+\sum_{k=2}^{\infty} a_kx^k\\ &=5+10x+\sum_{k=2}^{\infty} (5a_{k-1}-6a_{k-2}+2)x^k\\ &=5+10x+5\sum_{k=2}^{\infty} a_{k-1}x^k-6\sum_{k=2}^{\infty} a_{k-2}x^k+\sum_{k=2}^{\infty} 2x^k\\ &=5+10x-5a_0x+5x\sum_{k=1}^{\infty} a_{k-1}x^{k-1}-6x^2\sum_{k=2}^{\infty} a_{k-2}x^{k-2}+\dfrac{2x^2}{1-x}\\ &=5-15x+5xA(x)-6x^2A(x)+\dfrac{2x^2}{1-x} \end{aligned}$

From which we get

$A(x)=\dfrac{5-15x+\dfrac{2x^2}{1-x}}{6x^2-5x+1}$

By partial fractions and Taylor series of $\dfrac{1}{1-x}$

$\begin{aligned} A(x)&=\dfrac{3}{1-2x}+\dfrac{1}{1-3x}+\dfrac{1}{1-x}\\ &=\sum_{k=0}^{\infty} (3\cdot 2^k+3^k+1)x^k \end{aligned}$

$\Longrightarrow a_n=3\cdot 2^n+3^n+1$

By substitution in $(1)$ we can get the closed form of $b_n$

$\begin{aligned} b_n&=c_{n-1}-c_{n-2}-3a_{n-2}+2^{n-2}\\ &=2-2^{n-3}-(2-2^{n-4})-3(3\cdot 2^{n-2}+3^{n-2}+1)+2^{n-2}\\ &=-33\cdot 2^{n-4}-3^{n-1}-3 \end{aligned}$

To summarize

$\begin{cases} a_n=3\cdot 2^n+3^n+1\\ b_n=-33\cdot 2^{n-4}-3^{n-1}-3\\ c_n=2-2^{n-2} \end{cases}$

Thus

$\begin{aligned} a_7+b_5c_6&=3\cdot 2^7+3^7+1+(-33\cdot 2^{5-4}-3^{5-1}-3)(2-2^{6-2})\\ &=2572+(-150)(-14)=\boxed{4672} \end{aligned}$