1. Sequences and series question.

Hi guys, rhe following is a question that I busted my brains over, but could not get, so help would be greatly appreciated.

And I don't know how to properly use latex, so please bear with me.

Q) A sequence satisfies $\displaystyle T_{n}\=\frac{\1}{2}(T_{n-1}\+\T_{n+1})$ with $\displaystyle T_{1}$ = 3 and $\displaystyle T_{2}$ = 2.

Find $\displaystyle T_{3}$ and $\displaystyle T_{4}$

Also, T1=2 and T2=7

Question without latex: A sequence satisfies [term 'n' = 0.5*[term 'n-1' + term 'n+1']. Find term's 3 and 4.

Sorry I have absolutely no idea how to do this, one, also my latex skills are pretty poor, so ANY HELP WOULD BE GREATLY APPRECIATED.

2. $\displaystyle T_{n}=\frac{1}{2}(T_{n-1}+T_{n+1})$ with $\displaystyle T_{1} = 3$ and $\displaystyle T_{2} = 2$ .

When n = 2, the problem becomes

$\displaystyle T_{2}=\frac{1}{2}(T_{1}+T_{3})$

Substitute the known values and find T3.

3. End the LaTex with /math, not \math.

4. The difference equation written in 'standard form' is...

$\displaystyle t_{n} - 2\ t_{n-1} + t_{n-2}=0$ , $\displaystyle t_{0}= 3\ ,\ t_{1}=2$ (1)

... and its general solution depends from the root of algebraic equation...

$\displaystyle x^{2} - 2\ x +1=0$ (2)

... i.e. x=1 with multeplicity 2. The general solution is then...

$\displaystyle t_{n}= c_{1} + n\ c_{2}$ (3)

... that with the 'initial conditions' becomes...

$\displaystyle t_{n} = 3 - n$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Hello, SeaNanners

$\displaystyle \text{A sequence satisfies: }\;T_{n}\:=\:\frac{1}{2}(T_{n-1}+T_{n+1})\;\text{ with }\,T_{1} = 3\,\text{ and }\,T_{2} = 2$

$\displaystyle \text{Find }\,T_{3}\,\text{ and }\,T_{4}$

Each term is the average of the term before and the term after.

We have: .$\displaystyle \begin{array}{|c|c|c|c|c} T_1 & T_2 & T_3 & T_4 & \hdots \\ \hline 3 & 2 & ? & \end{array}$

$\displaystyle \,T_2$ is the average of $\displaystyle \,T_1$ and $\displaystyle T_3.$

. . $\displaystyle 2 \:=\:\dfrac{3+T_3}{2} \quad\Rightarrow\quad \boxed{T_3 \:=\:1}$

We have: .$\displaystyle \begin{array}{|c|c|c|c|c} T_1 & T_2 & T_3 & T_4 & \hdots \\ \hline 3 & 2 & 1 & ? \end{array}$

$\displaystyle T_3$ is the average of $\displaystyle T_2$ and $\displaystyle T_4.$

. . $\displaystyle 1 \:=\:\dfrac{2+T_4}{2} \quad\Rightarrow\quad \boxed{T_4 \:=\:0}$
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