Solving recurrence equation with generating indices of positive indices I don't know how to solve recurrence equation with positive indices like $$a_{n+2}  + 4a_{n+1}+
4a_n = 7$$ by generating functions.
How to solve such kind of problems.

Hint. You may just multiply both sides of the following relation by $x^{n+2}$ and summing it:
$$
a_{n+2}  + 4a_{n+1}+4a_n = 7 \tag1
$$ to get
$$
\sum_{n=0}^{\infty}a_{n+2}x^{n+2}  + 4x\sum_{n=0}^{\infty}a_{n+1}x^{n+1}+4x^2\sum_{n=0}^{\infty}a_n x^n= 7x^2\sum_{n=0}^{\infty}x^n
$$ or
$$
\sum_{n=2}^{\infty}a_{n}x^{n}  + 4x\sum_{n=1}^{\infty}a_{n}x^{n}+4x^2\sum_{n=0}^{\infty}a_n x^n= 7x^2\sum_{n=0}^{\infty}x^n \tag2
$$ equivalently, setting $\displaystyle f(x):=\sum_{n=0}^{\infty}a_{n}x^{n} $, you formally get 
$$
f(x)-a_0-a_1x+4x(f(x)-a_0)+4x^2f(x)=7x^2\frac{1}{1-x} 
$$ that is
$$
(2x+1)^2f(x)=\frac{7x^2}{1-x}+(4a_0+a_1)x+a_0
$$
$$
f(x)=\frac{7x^2}{(1-x)(2x+1)^2}+\frac{(4a_0+a_1)x+a_0}{(2x+1)^2} \tag3
$$ Then by partial fraction decomposition and power series expansion, you are able to identify coefficients of both sides of $(3)$.
