The sequence below has an 
n
th
n 
th
  term rule of the form 
a
n
2
+
b
n
+
c
an 
2
 +bn+c, where 
a
a, 
b
b and 
c
c are whole numbers.
Find the 
n
th
n 
th
  term rule for this sequence.

11
→
20
→
33
→
50
→
71
→
11→20→33→50→71→

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 \( n^2 \).

The sequence below has an nth term rule of the form an^2 + bn + c, where a, b and c are whole numbers. Find the nth term rule for this sequence: 11 → 20 → 33 → 50 → 71.

Learn how to find the nth term rule for the quadratic sequence 11, 20, 33, 50, 71. This solution demonstrates the use of differences to derive the formula T(n) = 2n^2 + 3n + 6. Suitable for students in Grades 8-10.

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