help me with this sequence and series problem

Question

I cannot understand how the n term tends to
1/3((x1)+2(x2))

Answer 1

How is $x_3$ defined? It is defined as the arithmetic mean between $x_1$ and $x_2$. And what is $x_4$? $x_4$ is the arithmetic mean of $x_3$ and $x_2$.

My suggestion: draw a line segment and call $x_1$ to the left ending, $x_2$ to the right one. $x_3$ is right in the middle of the segment. $x_4$ in the middle, between $x_3$ and $x_2$. $x_5$ in the middle between $x_3$ and $x_4$. And so on and so forth. How do you get to $x_3$?

$$x_3 = 5 - 1.5$$

because 1.5 is half the distance between $x_1$ and $x_2$. How do you get to $x_4$? $x_5$? $x_6$?

$$x_4 = x_3 + 0.75 = 5 - 1.5 + 0.75\\ x_5 = x_4 - 0.375 = 5 - 1.5 + 0.75 - 0.375\\ x_6 = x_5 + 0.1875 = 5 - 1.5 + 0.75 - 0.375 + 0.1875$$

if you keep doing this, you should be able to convince yourself that $x_n$ has an expression that starts with $5$ and then an alternating sum. Well, if you factor $1.5$ out, you get:

$$x_n = 5 + 1.5(-1 + 0.5 - 0.25 + 0.125 - \cdots )$$

But that alternating sum is

$$\sum_{i=0}^n (-1)^{i+1}\frac1{2^i}$$

which, when $n \to \infty$, yields $-\frac23$. Plugging in, we get

$$n \to \infty\ \Longrightarrow\ 5 - 1.5\cdot(\frac23) = 4$$

The attached formula can be derived if you think separately of the positive and negative terms of the summation above for $x_n$.

Answer 2

We have $x_1=2$ , $x_2=5$ and $$x_n=\frac12\left(x_{n-1}+x_{n-2}\right)$$ $$x_n-x_{n-1}=\left(\frac{-1}2\right)\left(x_{n-1}-x_{n-2}\right)$$ \begin{align} x_2-x_1 &=\left(\frac{-1}{2}\right)^03 \\ x_3-x_2&=\left(\frac{-1}{2}\right)^13 \\ x_4-x_3&=\left(\frac{-1}{2}\right)^23\\ x_5-x_4&=\left(\frac{-1}{2}\right)^33\\ &\cdots\cdots\cdots\cdots\\ &\cdots\cdots\cdots\cdots\\ x_n-x_{n-1}&=\left(\frac{-1}{2}\right)^{n-2}3\\ \end{align} on adding we have $x_n-x_1=3\left(1+(-1/2)+(-1/2)^2+\cdots(-1/2)^{n-2}\right)$ Now $$\displaystyle\lim_{n\to\infty}x_n=2+3\times\frac{1}{1+1/2}=4$$

Answer 3

You may have proved that somewhere. It comes from the usual technique to solving homogeneous linear recurrence relations. You assume $x_n=ar^n$, plug that into the recurrence, get the characteristic polynomial, here $r^2=\frac 12(r+1)$ and solve for $r$. You get two roots, here $r=1, -\frac 12$ The general solution is then $x_n=An^1+Bn^{-1/2}$ and you evaluate $A,B$ from the initial conditions. As the second term dies away to nothing as $n$ gets large, $x_n \to A$ for $n$ large. You find $A=\frac 13(x_1+2x_2)$ when you do this.