The expression for a_n which satisfies a_0=0, | vedclass

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(A) Given the recurrence relation $a_n = a_{n-1} + a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$. This is the characteristic equation $r^2 - r - 1 = 0$. The r

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$\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$

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(A) Given the recurrence relation $a_n = a_{n-1} + a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$. This is the characteristic equation $r^2 - r - 1 = 0$. The roots are $r = \frac{1 \pm \sqrt{5}}{2}$. Let $a_n = A\left(\frac{1+\sqrt{5}}{2}\right)^n + B\left(\frac{1-\sqrt{5}}{2}\right)^n$. Using $a_0 = 0$,we get $A + B = 0$,so $B = -A$. Using $a_1 = 1$,we get $A\left(\frac{1+\sqrt{5}}{2}\right) - A\left(\frac{1-\sqrt{5}}{2}\right) = 1$. $A\left(\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\right) = 1$ $\Rightarrow A\sqrt{5} = 1$ $\Rightarrow A = \frac{1}{\sqrt{5}}$. Thus,$B = -\frac{1}{\sqrt{5}}$. Substituting these values,$a_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$.

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

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