The $r^{th}$ term $U_r$ of a sequence $U_1, U_2, ...$ is given by $U_r = U_{r-1} + 8(r-1)$. If $U_2 = 9$, we need to find the values of $U_1, U_3, U_4$ and obtain an expression for $U_r$ in terms of $r$.

1. Problem Description

The $r^{th}$ term $U_r$ of a sequence $U_1, U_2, ...$ is given by $U_r = U_{r-1} + 8(r-1)$. If $U_2 = 9$, we need to find the values of $U_1, U_3, U_4$ and obtain an expression for $U_r$ in terms of $r$.

2. Solution Steps

First, we are given the recurrence relation $U_r = U_{r-1} + 8(r-1)$.

Since $U_2 = 9$, we can find $U_1$ by setting $r=2$:

$U_2 = U_1 + 8(2-1)$

$9 = U_1 + 8(1)$

$U_1 = 9 - 8 = 1$

Next, we can find $U_3$ by setting $r=3$:

$U_3 = U_2 + 8(3-1)$

$U_3 = 9 + 8(2)$

$U_3 = 9 + 16 = 25$

Then, we find $U_4$ by setting $r=4$:

$U_4 = U_3 + 8(4-1)$

$U_4 = 25 + 8(3)$

$U_4 = 25 + 24 = 49$

Now, we want to find an expression for $U_r$ in terms of $r$.

$U_r = U_{r-1} + 8(r-1)$

We can write out the first few terms to see if we can identify a pattern.

$U_1 = 1$

$U_2 = U_1 + 8(1) = 1 + 8 = 9$

$U_3 = U_2 + 8(2) = 9 + 16 = 25$

$U_4 = U_3 + 8(3) = 25 + 24 = 49$

$U_5 = U_4 + 8(4) = 49 + 32 = 81$

Notice that the terms are $1^2, 3^2, 5^2, 7^2, 9^2, ...$

So $U_r = (2r-1)^2$.

Let's prove this using induction.

Base case: $r=1$, $U_1 = (2(1)-1)^2 = 1^2 = 1$, which matches our value.

Assume $U_k = (2k-1)^2$ for some $k \ge 1$.

Then $U_{k+1} = U_k + 8(k+1-1) = (2k-1)^2 + 8k = 4k^2 - 4k + 1 + 8k = 4k^2 + 4k + 1 = (2k+1)^2 = (2(k+1)-1)^2$.

So the formula holds for $U_{k+1}$. By induction, $U_r = (2r-1)^2$ for all $r \ge 1$.

3. Final Answer

$U_1 = 1$, $U_3 = 25$, $U_4 = 49$

$U_r = (2r-1)^2$