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A: Given $a_{n}=2a_{n-1}-2a_{n-2}$

The characteristic equation will be $r^{2}=2r-2$

$r^{2}-2r+2=0$

$r=\frac{2\pm \sqrt{(-2)^{2}-4*1*2}}{2}$

$r=\frac{2\pm \sqrt{-4}}{2}$

$r=\frac{2\pm 2\sqrt{-1}}{2}$

$r=\frac{2\pm 2i}{2}$

$r=1\pm i$

 

B: $a_{0}=1\, \, \, a_{1}=2$

The solution of recurrence relation is of form $a_{n}=\alpha r_{1}^{n} + \beta r_{2}^n$

$a_{n}=\alpha (1-i)^{n} + \beta (1+i)^n$

at n=0 and n=1 we have

$1=\alpha + \beta$

$2=(1-i)\alpha + (1+i)\beta$

Multiply first equation by $(1-i)$

$(1-i)=(1-i)\alpha + (1-i)\beta$

$2=(1-i)\alpha + (1+i)\beta$

subtract two equations

$1+i=2i\beta$

$\beta =\frac{i+1}{2i}$

$\beta =\frac{(i+1)i}{2i^2}$

$\beta =\frac{i-1}{-2}$

$\beta =\left ( \frac{1}{2}-\frac{i}{2} \right )$

$\alpha =1-\beta$

$\alpha =1-\left ( \frac{1}{2}-\frac{i}{2} \right )$

$\alpha =\left ( \frac{1}{2}+\frac{i}{2} \right )$

Hence the solution becomes:

$a_{n} =\left ( \frac{1}{2}+\frac{i}{2} \right )\cdot (1-i)^{n} + \left ( \frac{1}{2}-\frac{i}{2} \right )\cdot (1+i)^n$

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