Finding an Expression for t(n)

[issoder]

hi all,

Would it be possible for you to explain how to reach a solution for this question, or explain the process that you need to find the solution
View attachment 2469

thank you

 

[chisigma]

hi all,

Would it be possible for you to explain how to reach a solution for this question, or explain the process that you need to find the solution
View attachment 2469

thank you

Wellcome on MHB issoder!...

The general procedure for solving a first order linear difference equation is illustrated here...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-i-426.html#post2494

This is one of simplest cases because the coefricients are constant. A difference equation in the form...

$\displaystyle t_{n+1} = \alpha\ t_{n} + \beta,\ t_{0}=a (1)$

... has solution...

$\displaystyle t_{n} = a\ \alpha^{n} + \beta\ \frac{1 - \alpha^{n}}{1 - \alpha}\ (2)$

In Your case is $\alpha= 2$, $\beta=4$ and $a=2$ so that is $\displaystyle t_{n} = 2^{n+2} + 2^{n+1} - 4$...

Kind regards

$\chi$ $\sigma$

 

[soroban]

Hello, issoder!

Let $$t_o,t_1,t_2,\cdots$$ be a sequence defined by:
. . $$t_0 \:=\:2$$
. . $$t_{n+1} \:=\:2t_n+4$$

Find an expression for $$t_n.$$

$$\begin{array}{cccccc}\text{We are given:} & t_{n+1} &=& 2t_n + 4 & [1] \\ \text{Next term:} & t_{n+2} &=& 2t_{n+1} + 4 & [2] \end{array}$$

$$\text{Subtract [2]-[1]: }\;t_{n+2} - t_{n+1} \;=\;2t_{n+1} - 2t_n$$

. . . . . . . $$t_{n+2} - 3t_{n+1} + 2t_n \;=\;0$$

Let $$X^n = t_n\!:\;\;X^{n+2} - 3X^{n+1} + 2X^n \;=\;0$$

Divide by $$X^n\!:\;\;X^2 - 3X + 2 \;=\;0$$

Then: .$$(X-1)(X-2) \:=\:0 \quad\Rightarrow\quad X \:=\:1,2$$

The function is: .$$f(n) \:=\: (1^n)A + (2^n)B$$

We know the first two terms: $$t_0 = 2,\;t_1 = 8$$

$$\begin{array}{cccccc}f(0) = 2: & A + B &=& 2 & [3] \\ f(1) = 8: & A + 2B &=& 8 & [4] \end{array}$$

Subtract [4]-[3]: .$$B = 6 \quad\Rightarrow\quad A = -4$$

Hence: .$$f(n) \;=\;-4(1^n) + 6(2^n)$$

Therefore: .\$$t_n \;=\;f(n) \;=\;6\!\cdot\!2^n - 4$$