$\displaystyle a_{n+1}=7a_n-6,$ and $\displaystyle a_1=10,a_2=64$
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.
$\displaystyle \text{Given: }\:a_{n+1}\:=\:7a_n-6,\;\text{ and }\,a_1=10,\;a_2=64$
. . $\displaystyle \text{Find the closed form: }f(n).$
By the way, the answer is:.$\displaystyle f(n) \:=\:9\cdot7^{n-1} + 1$
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.