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

• May 25th 2011, 09:31 AM
Melsi
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!
• May 25th 2011, 03:18 PM
Soroban
Hello, Melsi!

Quote:

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

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

$\displaystyle \text{Solution}$

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

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

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

$\displaystyle A_1 \;=\;a$

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

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

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

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

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

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

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

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

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

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

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

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

• May 26th 2011, 02:33 AM
Melsi
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!!
(Clapping)