# solve equation

• Nov 14th 2011, 11:30 AM
greg1987
solve equation
$a_{n+1}=7a_n-6,$ and $a_1=10,a_2=64$
• Nov 14th 2011, 11:38 AM
Quacky
Re: 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 AM
greg1987
Re: solve equation
IN this case we don't want recurrence relationship, we need determine $a_n$ using $n$ only.
• Nov 14th 2011, 02:15 PM
Soroban
Re: solve equation
Hello, greg1987!

Have you been taught anything about recurrence relations?

It's impossible to help you if I don't know what you have learned so far
. . and where you are having difficulty.

Quote:

$\text{Given: }\:a_{n+1}\:=\:7a_n-6,\;\text{ and }\,a_1=10,\;a_2=64$
. . $\text{Find the closed form: }f(n).$

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

• Nov 15th 2011, 12:31 AM
sbhatnagar
Re: solve equation
Quote:

Originally Posted by greg1987
$a_{n+1}=7a_n-6,$ and $a_1=10,a_2=64$

Is $a_n$ the $n^{th}$ term of an arithmetic progression?
• Nov 15th 2011, 01:19 AM
mr fantastic
Re: solve equation
Quote:

Originally Posted by sbhatnagar
Is $a_n$ the $n^{th}$ term of an arithmetic progression?

Clearly not, since the recurrence relation would then take the form $a_{n+1} - a_n = d$ where d is a constant, and obviously this is not the case. Not to mention the fact that post #4 obviously answers your question in the negative.
• Nov 17th 2011, 04:48 AM
HallsofIvy
Re: solve equation
For "linear recurrences with constant coefficients", things such as $a_{n+1}= ba_n+ c$ where b and c are numbers, there is a very general method: first drop the number that is not multiplying and "a": $a_{n+1}= ba_n$ and "try" $a_n= r^n$. Then $a_{n+1}= r^{n+1}$ so the equation becomes $r^{n+1}= br^n$ and, dividing both sides by $r^n$, r= b. That is $a_n= b^n$ satisfies $a_{n+1}= ba_n$. In fact it is easy to see that any constant times that, $a_n= Pb^n$, is a solution. To find a solution to the entire equation, "try" a constant: if $a_n= A$ the equation becomes $A= bA+ c$ so $(1- b)A= c$ and (if b is not 1), $A= \frac{c}{1- b}$. The solution to the entire equation is the sum of those: $a_n= Pb^n+ \frac{c}{1- b}$ for any constant p. (If b= 1, the equation is $a_{n+1}= a_n+ c$ and we have an arithmetic sequence.)

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