# sequence formula

• Oct 28th 2007, 03:44 AM
slevvio
sequence formula
The sequence {bn} is defined by b1 = 2, b2 = 8, b3 = 8bn-1 - 7bn-2 (n ≥ 3). Find b3 and b4 and guess a formula for bn. Use induction to prove your conjecture.

The first part of this is easy, b3 and b4 being 50 and 344 respectively, but what I am having trouble with is guessing a general formula and I was wondering if anybody could help me. I know how to prove it but I don't have anything to prove yet! Thanks a lot:)
• Oct 28th 2007, 06:46 AM
Plato
Hint: $\displaystyle 50 = 1 + 7^2$
• Oct 28th 2007, 06:47 AM
Soroban
Hello, slevvio!

Quote:

The sequence $\displaystyle \{b_n\}$ is defined by: .$\displaystyle b_1 \:= \:2,\;b_2 \:= \:8,\;b_3 \:=\:8b_{n-1} - 7b_{n-2},\;\; (n \geq 3)$

Find $\displaystyle b_3$ and $\displaystyle b_4$ and guess a formula for $\displaystyle b_n.$
Use induction to prove your conjecture.

There is an intricate algebraic procedure for determining the general term.
But as I started it, I eyeballed the first few terms ... and saw a pattern!

The first few terms are:

. . $\displaystyle \begin{array}{ccc}b_1 & = & 2 \\ b_2 & = & 8 \\ b_3 & = & 50 \\ b_4 & = & 344 \\ b_5 & = & 2402 \\ \vdots & & \vdots\end{array}$

Those values looked "kind of familiar" . . .
Then I recalled the powers-of-7 (which I had memorized years ago):
. . $\displaystyle \begin{array}{ccc}7^0 & = & 1 \\ 7^1 & = & 7 \\ 7^2 & = & 49 \\ 7^3 & = & 343 \\ 7^4 & = & 2401 \\ \vdots & & \vdots\end{array}$

Get it? . . . Those terms are one more than a power-of-7.

I think we've got it! . . . $\displaystyle \boxed{b_n \;=\;7^{n-1} + 1}$

I tested my conjecture with the next term: .$\displaystyle b_5\:=\:8(2402) - 7(344) \:=\:16808$

. . which just happens to be: .$\displaystyle 7^5 + 1$ . . . Yay!