How do i find the n-th term of a sequence, for which $a_{0}=0$, $a_{1}=1$, $a_{n+2}=\dfrac{a_{n+1}+a_{n}}{2}$
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Try using the same trick used for the Fibonacci sequence – QuantumFool Oct 23 '16 at 20:53
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I guess you mean $a_0=0,a_1=1$ instead of $a_1=0, a_1=1$? – Inzinity Oct 23 '16 at 20:55
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This is linear homogenous reccurence relation and it's characteristic equation is: $2x^2 - x - 1 = 0$. It's solutions are $x=1$ and $x= - \frac{1}{2}$. Hence we have that:
$$a_n = A\cdot 1^n + B \cdot \left(-\frac 12\right)^n$$
Now plug in the values for $n=0,1$ and solve the system of linear equations to get the values of $A$ and $B$.
Stefan4024
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A standard way, for linear difference equations, is to consider the form $a_{n} = r^{n}$ for which the difference equation $2 \, a_{n+2} = a_{n+1} + a_{n}$ leads to the quadratic equation $2 r^2 - r - 1 = 0$. The solutions are $r \in \{ 1, -\frac{1}{2} \}$. The form of the solution to the difference equation is then $$a_{n} = c_{0} + c_{1} \, \frac{(-1)^{n}}{2^{n}}.$$ From the condition $a_{0} = 0$ then $c_{1} = - c_{0}$ and $$a_{n} = c_{0} \, \left( 1 - \frac{(-1)^{n}}{2^{n}} \right)$$ and from $a_{1} = 1$ then $3 c_{0} = 2$ for which $$a_{n} = \frac{2}{3} \, \left( 1 - \frac{(-1)^{n}}{2^{n}} \right)$$
Leucippus
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