Thread: Finding General Term Of Sequence (unable to understand solved exercise on book)

1. Finding General Term Of Sequence (unable to understand solved exercise on book)

Hello,

Given the sequence (An) with An+1 = BAn + C , A1=a, C not zero, B not 1

Find the formula (for general term) of this sequence.

Solution

A2 = BA1 + C = BA + C ((1-B)/(1-B))
A3=...=(B^2)a + C((1-B^2)/(1-B))
.... etc
it continues, then uses math induction etc and gives the solution that An = (B^n-1)M+N

I am not really asking for the result, my problem is very specific, is this:

((1-B)/(1-B))... which I don't understand where came from.

Thank you all!

2. Hello, Melsi!

$\displaystyle \text{Given the sequence }\{A_n\}\text{ with }A_{n+1} \,=\,BA_n + C$
. . $\displaystyle A_1=a,\; C \ne 0,\; B \ne 1$

$\displaystyle \text{Find the formula for general term of this sequence.}$

$\displaystyle \text{Solution}$

$\displaystyle A_2 \;=\; BA_1 + C \;=\; Ba + C\left(\frac{1-B}{1-B}\right)$

$\displaystyle A_3 \;=\;\hdots\;=\; B^2a + C\left(\frac{1-B^2}{1-B}\right) \quad \hdots\text{ etc.}$

$\displaystyle \text{It continues, then uses math induction, etc.}$
$\displaystyle \text{and gives the solution: }\:A_n \:=\: B^{n-1}M+N$ . Really?

$\displaystyle \text{I am not asking for the result. \;My problem is very specific:}$
. . $\displaystyle \frac{1-B}{1-B}\;\hdots\text{ I don't understand where it came from.}$

They found a pattern in the successive terms and used it.

$\displaystyle A_1 \;=\;a$

$\displaystyle A_2 \;=\;B(a) + C \;=\;Ba + C$

$\displaystyle A_3 \;=\;B(Ba + C) + C \;=\;B^2a + BC + C \;=\;B^2a + C(B+1)$

$\displaystyle A_4 \;=\;B[B^2a + C(B+1)]+C \;=\;B^3a + C(B^2 + B + 1)$

$\displaystyle A_5 \;=\;B[B^3a + C(B^2+B+1)] \;=\;B^4a + C(B^3 + B^2 + B + 1)$

. . . . $\displaystyle \hdots \text{ etc.}$

$\displaystyle \text{We note that those final polynomials can be be rewritten:}$

. . $\displaystyle A_5 \;=\;B^4a + C(B^3 + B^2 + B + 1) \;=\;B^4a + C\left(\frac{B^4 - 1}{B - 1}\right)$

. . $\displaystyle A_4 \;=\;B^3a + C(B^2 + B + 1) \;=\;B^3a + C\left(\frac{B^3 - 1}{B - 1}\right)$

. . $\displaystyle A_3 \;=\; B^2a + C(B + 1) \;=\;B^2a + C\left(\frac{B^2-1}{B-1}\right)$

$\displaystyle \text{And to complete the pattern, they wrote }A_2\text{ like this:}$

. . $\displaystyle A_2 \;=\;Ba + C \;=\;Ba + C\left(\frac{B-1}{B-1}\right)$

$\displaystyle \text{The formula is: }\;A_n \;=\;B^{n-1}a + C\left(\frac{B^{n-1}-1}{B-1}\right)$

3. Yes you are right!

I didn't see the pattern, in every line the Ba+C is present and the only thing that changes is the power of B and the coefficient of C.

Unfortunately my reasoning was way more complicated including equations and roots in order to transform it to a geometric sequence which led me in nowhere!

It is a special sequence used in economy, and I guess it has its own characteristics and in order to make it geometric or arithmetic you would have to set B or C to special values (but then the general case is lost and and specific ones come up).

Thank you very very much, it was the piece I missed from the puzzle
I will keep in mind this pattern trick!!