# Thread: Conjuncture a Formula and Probe it Using Induction

1. ## Conjuncture a Formula and Probe it Using Induction

Hey all. I have a problem in my stage 2 pure maths paper that I'm not sure how to solve. Here it is:

Let x(1), x(2), x(3),... be a sequence of integers defined recursively by :
x(1) = 3, x(2) = 18, and x(n) = 6x(n-1) - 9x(n-2)
for integers n >= 3. Conjecture a formula for x(n) and prove it using mathematical induction.

I'm really not sure how to start it... I think I will be fine with the induction but am not sure how to get the formula...

2. Hello, mrtwigx!

Finding the formula can be quite a task . . . or quite simple.

Let $\displaystyle x(1),\:x(2),\:x(3)\:\hdots$ be a sequence of integers defined recursively by:

. . $\displaystyle x(1) = 3,\;\;x(2) = 18,\;\;x(n) \:=\: 6\!\cdot\!x(n-1) - 9\!\cdot\!x(n-2)\;\text{ for }n \geq 3$

Conjecture a formula for $\displaystyle x(n)$ and prove it using mathematical induction.
Crank out the first few terms of the sequence and hope for a pattern:

. . $\displaystyle \begin{Bmatrix}x(1) &=& 3 &=& 1\cdot 3\\ x(2) &=& 18 &=& 2\cdot3^2\\ x(3) &=& 81 &=& 3\cdot3^2\\ x(4) &=& 324 &=& 4\cdot 3^4 \\x(5) &=& 1215 &=& 5\cdot3^5 \end{Bmatrix}\qquad \Leftarrow\:\text{ There!}$

The $\displaystyle n^{th}$ term seems to be: .$\displaystyle x(n) \;=\;n\!\cdot\!3^n$

Go for it!

3. Thanks