To find the $n$-th term rule for the sequence:

$11, 20, 33, 50, 71, \dots$

We assume the $n$-th term rule is of the form:

$T(n) = an^2 + bn + c$

Step 1: Find the differences between the terms.

First, compute the first differences (difference between consecutive terms):

$20 - 11 = 9$ $33 - 20 = 13$ $50 - 33 = 17$ $71 - 50 = 21$

So, the first differences are: $9, 13, 17, 21$.

Now compute the second differences (differences of the first differences):

$13 - 9 = 4$ $17 - 13 = 4$ $21 - 17 = 4$

Since the second differences are constant, this confirms the $n$-th term rule is quadratic.

Step 2: Use the quadratic form $T(n) = an^2 + bn + c$.

We know that the second difference being constant means the coefficient of $n^2$ is half of this second difference. Since the second difference is $4$, we have:

$a = \frac{4}{2} = 2$

So, the rule starts as $T(n) = 2n^2 + bn + c$.

Step 3: Find $b$ and $c$.

Now, substitute values from the sequence to solve for $b$ and $c$.

When $n = 1$:

$T(1) = 2(1)^2 + b(1) + c = 11$ $2 + b + c = 11 \quad \Rightarrow \quad b + c = 9 \quad \text{(Equation 1)}$

When $n = 2$:

$T(2) = 2(2)^2 + b(2) + c = 20$ $8 + 2b + c = 20$ $2b + c = 12 \quad \text{(Equation 2)}$

Step 4: Solve the system of equations.

From Equation 1: $b + c = 9$
From Equation 2: $2b + c = 12$

Subtract Equation 1 from Equation 2:

$(2b + c) - (b + c) = 12 - 9$ $b = 3$

Now substitute $b = 3$ into Equation 1:

$3 + c = 9 \quad \Rightarrow \quad c = 6$

Step 5: Write the $n$-th term rule.

Thus, the $n$-th term rule is:

$T(n) = 2n^2 + 3n + 6$

Verify the rule with the sequence:

$T(1) = 2(1)^2 + 3(1) + 6 = 11$ $T(2) = 2(2)^2 + 3(2) + 6 = 20$ $T(3) = 2(3)^2 + 3(3) + 6 = 33$ $T(4) = 2(4)^2 + 3(4) + 6 = 50$ $T(5) = 2(5)^2 + 3(5) + 6 = 71$

The formula works!

Would you like further details on this?

Here are 5 related questions to deepen your understanding:

1. How do you determine the degree of a sequence from its differences?
2. What happens if the second differences are not constant?
3. How would the process change if the sequence was cubic?
4. Can you derive a sequence with negative second differences?
5. How does changing the value of $a$ affect the sequence?

Tip: Quadratic sequences always have constant second differences, and you can use these differences to quickly find the coefficient of $n^2$.