How Can You Find 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