# A sequence is defined by a_n=a_(n-1)+2a_(n-2), with a_1=-3, a_2=4. What is the value of a_6?

Jan 12, 2017

${a}_{6} = 14$. The general formulas ${a}_{n} = {\left(- 1\right)}^{n} \left(\frac{10}{3}\right) + \frac{1}{6} {2}^{n}$ gives

${a}_{10} = \frac{10}{3} + \frac{1024}{6} = 174$.

#### Explanation:

Define the shift operator E by

$E {a}_{n - 2} = {a}_{n - 1}$

Now, the given equation becomes

$f \left(E\right) {a}_{n} = \left({E}^{2} - E - 2\right) {a}_{n} = 0 , n = 1 , 2 , 3 , \ldots$

The general solution of this difference equation is

${a}_{n} = A {\left({\rho}_{1}\right)}^{n} + B {\left({\rho}_{2}\right)}^{n}$, where ${\rho}_{1} \mathmr{and} {\rho}_{2}$ are the

zeros of

$f \left(\rho\right) = {\rho}^{2} - \rho - 2 = 0$, giving $\rho = 1 \mathmr{and} 2.$

Now,

${a}_{n} = {\left(- 1\right)}^{n} A + {2}^{n} B$

The initial value ${a}_{1} = - 3$ gives

$- A + 2 B = - 3$

and ${a}_{2} = 4$ gives

$A + 4 B = 4$.

Solving,

$A = \frac{10}{3} \mathmr{and} B = \frac{1}{6}$

It follows that

${a}_{6} = \frac{10}{3} + \left({2}^{6}\right) \left(\frac{1}{6}\right) = 14$.