I don't know what methods are available to you, so I'll just use one that I'm comfortable with: generating functions. It's a bit tedious, but it works! If you don't know it, there's no harm in learning about it.
Let U(x) be the generating function for the sequence u(n), i.e.
[tex]\displaystyle U(x) = \sum_{n=0}^\infty u(n)x^n[/tex]
In the recurrence equation, we multiply both sides by xⁿ (where |x| < 1, which will come into play later), then take the sums on both sides from n = 0 to ∞, thus recasting the equation as
[tex]\displaystyle \sum_{n=0}^\infty u(n+2) x^n = 8 \sum_{n=0}^\infty u(n+1) x^n - 16 \sum_{n=0}^\infty u(n) x^n[/tex]
Next, we rewrite each sum in terms of U(x). For instance,
[tex]\displaystyle \sum_{n=0}^\infty u(n+2) x^n = \frac1{x^2} \sum_{n=0}^\infty u(n+2) x^{n+2} \\\\ \sum_{n=0}^\infty u(n+2) x^n = \frac1{x^2} \bigg(u(2)x^2 + u(3)x^3 + u(4)x^4 + \cdots \bigg) \\\\ \sum_{n=0}^\infty u(n+2) x^n = \frac1{x^2} \sum_{n=2}^\infty u(n) x^n \\\\ \sum_{n=0}^\infty u(n+2) x^n = \frac1{x^2} \left(\sum_{n=0}^\infty u(n) x^n - u(1)x - u(0)\right) \\\\ \sum_{n=0}^\infty u(n+2) x^n = \frac1{x^2}(U(x) - 16x - 1) \\\\ \sum_{n=0}^\infty u(n+2) x^n = \frac1{x^2}U(x) - \frac{16}x - \frac1{x^2}[/tex]
After rewriting each sum in a similar way, we end up with a linear equation in U(x),
[tex]\displaystyle \frac1{x^2}U(x) - \frac{16}x - \frac1{x^2} = \frac8x U(x) - \frac8x - 16 U(x)[/tex]
Solve for U(x) :
[tex]\displaystyle \left(\frac1{x^2}-\frac8x+16\right) U(x) = \frac1{x^2} + \frac8x \\\\ \left(1-8x+16x^2\right) U(x) = 1 + 8x \\\\ (1-4x)^2 U(x) = 1 + 8x \\\\ U(x) = \dfrac{1+8x}{(1-4x)^2}[/tex]
The next step is to get the power series expansion of U(x) so that we can easily identity u(n) as the coefficient of the n-th term in the expansion.
Recall that for |x| < 1, we have
[tex]\displaystyle \frac1{1-x} = \sum_{n=0}^\infty x^n[/tex]
By differentiating both sides, we get
[tex]\displaystyle \frac1{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1} = \sum_{n=0}^\infty (n+1)x^n[/tex]
It follows that
[tex]\displaystyle \frac1{(1-4x)^2} = \sum_{n=0}^\infty (n+1)(4x)^n[/tex]
and so
[tex]\displaystyle \frac{1+8x}{(1-4x)^2} = \sum_{n=0}^\infty (n+1)(4x)^n + 8x\sum_{n=0}^\infty (n+1)(4x)^n \\\\ \frac{1+8x}{(1-4x)^2} = \sum_{n=0}^\infty 4^n(n+1)x^n + 2\sum_{n=0}^\infty 4^{n+1}(n+1)x^{n+1} \\\\ \frac{1+8x}{(1-4x)^2} = \sum_{n=0}^\infty 4^n(n+1)x^n + 2\sum_{n=1}^\infty 4^nnx^n \\\\ \frac{1+8x}{(1-4x)^2} = \sum_{n=0}^\infty 4^n(n+1)x^n + 2\sum_{n=0}^\infty 4^nnx^n \\\\ \frac{1+8x}{(1-4x)^2} = \sum_{n=0}^\infty 4^n(3n+1)x^n[/tex]
which means
[tex]u(n) = \boxed{4^n(3n+1)}[/tex]
