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The first two terms of a sequence are 

a1 = 2 and 

a2 = 7

After that ever term is half of the sum of previous two terms. Thus 

a3 = (a2 + a1)/2 

a4 = (a3 + a2)/2

and so on. How can I find the non-recursive formula for nth term?

Key Flex
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Avinash
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    Stock answer is to find the roots of the characteristic polynomial $2t^2-t-1=0$, then choose the constants as to match the initial conditions. If this doesn't make sense, please post (a lot) more context about what you do know about recurrences, which you are supposed to use here. Also see How to ask a good question. – dxiv Jul 18 '18 at 04:04
  • In linear recurrences like this, the first step is to find solutions of the form $a_n = r^n$, ignoring the initial conditions. In this instance, you will find two distinct values of $r$ that work, call them $r_1$ and $r_2$. Then there is a theorem that says that every solution is of the form $a_n = \alpha_1 r_1^n + \alpha_2 r_2^n$ for some constant coefficients $\alpha_1, \alpha_2$. The correct coefficients can be found by using the initial conditions. – Dave Jul 18 '18 at 04:13
  • Related https://math.stackexchange.com/questions/2559687/show-that-the-sequence-a-1-1-a-2-2-a-n2-a-n1a-n-2-converges-b/2559826#2559826 – rtybase Jul 18 '18 at 07:39

4 Answers4

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I give you a hint, observe that we can write $(a_n,a_{n-1}) = A(a_{n-1},a_{n-2})$ with A a 2x2 matrix.

Now $(a_n,a_{n-1}) = A^{n-1} \times (a_1,a_0)$. So we change the problem of finding a recursive formula to elevate a matrix n-1 times. To do the last, you can study the eignvalues of $A$.

Key Flex
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HFKy
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Assuming $a_n=a^n$ implies $$ a^{n+1} = \frac {a^n +a^{n-1}}{2}$$ We solve the equation $$ a^2 = \frac {a +1}{2}$$ or $$2a^2-a-1 =0$$

After solving for $a$ and finding two values $x_1$ and $x_2$ the general form of solution is$$ a_n = Ax_1^n + Bx_2 ^n$$ where $A$ and $B$ are found from the initial conditions.

Mohammad Riazi-Kermani
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  • @Dave You are correct. That was very bad mistake. I edited my solution – Mohammad Riazi-Kermani Jul 18 '18 at 04:20
  • Thanks to all of you. With your help I could solve the problem. I got a(n) = (20/3) (-1/2)^n + (16/3) – Avinash Jul 18 '18 at 04:53
  • Are we allowed to start from a^(n+1) = [a^n + a^(n-1)]/2 ? Using the suggestion by Alex.B I found the eigenvalues of the transformation matrix and found them to be (-1/2) and 1. Then I put x1 = -1/2 and x2 = 1 in the answer provided by Mohammad Riazi-Kermani and determined the constants A and B from initial conditions. Is this approach correct? – Avinash Jul 18 '18 at 05:01
  • One you found the eignvalues of the matrix you can find A^n (a1,a0) as follows: (a1,a0) = x 1v1 + x2v2 where v1 and v2 are the eignvectors and x1 and x2 the coordinates. Hence A^n (a1,a0) = A^n (x1v1 + x2v2) = x1b1^nv1 + x2b2^nv2, where b1 and b2 are the eignvalues you found. Finally (an,an-1)= x1b1^nv1 + x2b2^nv2. Then you have an. – HFKy Jul 18 '18 at 13:35
  • Another option is calculate A^n, A^n = PD^nP^-1, where D is the diagonal matrix and P the change of coordinates matrix. This approach is completely equivalent to that given by the other users. ( At least I believe this) – HFKy Jul 18 '18 at 13:37
  • Observe that even what I suggest could be a proof of the procedure described by Mostafa – HFKy Jul 18 '18 at 13:40
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$$a_{n+1}=\frac{1}{2}(a_n+a_{n-1})$$ could be rewritten as $$a_{n+1}-a_n=-\frac{1}{2}(a_n-a_{n-1}),$$

which shows that $\{a_n-a_{n-1}\}$ is a geometric sequence. Thus, $$a_n-a_{n-1}=(a_2-a_1)\cdot \left(-\frac{1}{2}\right)^{n-2}=5 \cdot \left(-\frac{1}{2}\right)^{n-2},$$where $n=2,3,\cdots.$

It follows that

$$\sum_{k=2}^n \left(a_k-a_{k-1}\right)=\sum_{k=2}^n \left[5 \cdot \left(-\frac{1}{2}\right)^{k-2}\right],$$

namely, $$a_n-a_1=5\sum_{k=2}^n\left(-\frac{1}{2}\right)^{k-2}. $$ As a result, $$a_n=5\sum_{k=2}^n\left(-\frac{1}{2}\right)^{k-2}+a_1=\frac{20}{3}\left(-\frac{1}{2}\right)^n+\frac{16}{3},$$ which also holds for $n=1.$

mengdie1982
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The characteristic equation is $$2u^2=u+1$$whose roots are $1,-\dfrac{1}{2}$therefore $$a_n=A+B\left(-\dfrac{1}{2}\right)^n$$from $a_1=2$ and $a_2=7$ we obtain$$a_n=-\dfrac{4}{3}+\dfrac{20}{3}\left(-\dfrac{1}{2}\right)^n$$

Mostafa Ayaz
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