Equation: $a_n = a_{n-1}-n$ for the given condition: $a_0 = 4$
So that means I know that $a_1 = 3$, $a_2 = 1$, $a_3 = -2$, $a_4 = -6$, and $a_5 = -11$.
I've tried for over 1.5 hours now trying to solve this. Here's what I've had.
$a_1 = 4 - 1$
$a_2 = 4 - 3(n-1)$
$a_3 = 4 - 3(n-1) - 0$
$a_4 = 4 - 3(n-1) - 1$
$a_5 = 4 - 3(n-1) - 3$
I'm not sure what I'm doing now. I'm supposed to find a pattern that repeats and forms an equation that works for all the values when plugged in back into the original equation by replacing $a_{n-1}$.
By pattern, here's what I mean. This is a problem that I already solved. I had the following equation: $a_n = -a_{n-1}$ for $a_0 = 5$. Here was the pattern I found for the values $a_1 = -5$, $a_2 = 5$, $a_3 = -5$, $a_4 = 5$ and so on. Here's the pattern:
$a_1 = -(5)$
$a_2 = -(-(5))$
$a_3 = -(-(-(5)))$
So the pattern is $a_n = -5(-1)^n$ because when you plug in the value of n, it works for all of them for the equation $a_n = -(-5(-1)^n)$.
Likewise, this problem $a_n = a_{n-1}+3$ for $a_0 = 1$ had the solution $a_n = 3(n-1)+1$ and plugging into the original $a_n = 3(n-1)+1+3$ results in $a_n = 3n+1$ which is true for its values {1, 4, 7, 10,...}.