$\displaystyle a_{n+1}=7a_n-6,$ and $\displaystyle a_1=10,a_2=64$

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- Nov 14th 2011, 11:30 AMgreg1987solve equation
$\displaystyle a_{n+1}=7a_n-6,$ and $\displaystyle a_1=10,a_2=64$

- Nov 14th 2011, 11:38 AMQuackyRe: solve equation
What do you mean "solve equation"? There isn't anything to solve. You have a recurrence relation. What is the exact wording of the question?

- Nov 14th 2011, 11:44 AMgreg1987Re: solve equation
IN this case we don't want recurrence relationship, we need determine $\displaystyle a_n$ using $\displaystyle n$ only.

- Nov 14th 2011, 02:15 PMSorobanRe: solve equation
Hello, greg1987!

Have you been taughtabout recurrence relations?*anything*

It's impossible to help you if I don't know what you have learned so far

. . and where you are having difficulty.

Quote:

$\displaystyle \text{Given: }\:a_{n+1}\:=\:7a_n-6,\;\text{ and }\,a_1=10,\;a_2=64$

. . $\displaystyle \text{Find the closed form: }f(n).$

By the way, the answer is:.$\displaystyle f(n) \:=\:9\cdot7^{n-1} + 1$

- Nov 15th 2011, 12:31 AMsbhatnagarRe: solve equation
- Nov 15th 2011, 01:19 AMmr fantasticRe: solve equation
- Nov 17th 2011, 04:48 AMHallsofIvyRe: solve equation
For "linear recurrences with constant coefficients", things such as $\displaystyle a_{n+1}= ba_n+ c$ where b and c are numbers, there is a very general method: first drop the number that is not multiplying and "a": $\displaystyle a_{n+1}= ba_n$ and "try" $\displaystyle a_n= r^n$. Then $\displaystyle a_{n+1}= r^{n+1}$ so the equation becomes $\displaystyle r^{n+1}= br^n$ and, dividing both sides by $\displaystyle r^n$, r= b. That is $\displaystyle a_n= b^n$ satisfies $\displaystyle a_{n+1}= ba_n$. In fact it is easy to see that any constant times that, $\displaystyle a_n= Pb^n$, is a solution. To find a solution to the entire equation, "try" a constant: if $\displaystyle a_n= A$ the equation becomes $\displaystyle A= bA+ c$ so $\displaystyle (1- b)A= c$ and (if b is not 1), $\displaystyle A= \frac{c}{1- b}$. The solution to the entire equation is the sum of those: $\displaystyle a_n= Pb^n+ \frac{c}{1- b}$ for any constant p. (If b= 1, the equation is $\displaystyle a_{n+1}= a_n+ c$ and we have an arithmetic sequence.)

This same idea can be used for higher order recurrences: If we have $\displaystyle a_{n+2}= 3a_{n+1}- 2a_n$, setting $\displaystyle a_n= r^n$ gives [tex]r^{n+2}= 3r^{n+1}- 2r^n[tex] and dividing through by $\displaystyle r^n$ and moving everything to the left, $\displaystyle r^2- 3r+ 2= (r- 1)(r- 2)= 0$. That is, both $\displaystyle a_n= 1^n= 1$ and $\displaystyle a_n= 2^n$ satisfy the recurrence and the general solution is $\displaystyle A(1^n)+ B(2^n)= A+ B2^n$ for any constants, A and B.