Cubic Sequences worksheet - Free Printable

Problem Analysis:

The task involves finding the formula for the \( n \)-th term of two cubic sequences. A cubic sequence is one where the third differences between consecutive terms are constant. We will solve each exercise step by step.

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Exercise 1:

The sequence is given as:
\[ 4, 14, 40, 88, 164, \ldots \]

#### Step 1: Calculate the first differences.
The first differences are obtained by subtracting consecutive terms:
\[
\begin{aligned}
&14 - 4 = 10, \\
&40 - 14 = 26, \\
&88 - 40 = 48, \\
&164 - 88 = 76.
\end{aligned}
\]
So, the first differences are:
\[ 10, 26, 48, 76, \ldots \]

#### Step 2: Calculate the second differences.
The second differences are obtained by subtracting consecutive first differences:
\[
\begin{aligned}
&26 - 10 = 16, \\
&48 - 26 = 22, \\
&76 - 48 = 28.
\end{aligned}
\]
So, the second differences are:
\[ 16, 22, 28, \ldots \]

#### Step 3: Calculate the third differences.
The third differences are obtained by subtracting consecutive second differences:
\[
\begin{aligned}
&22 - 16 = 6, \\
&28 - 22 = 6.
\end{aligned}
\]
So, the third differences are:
\[ 6, 6, \ldots \]

Since the third differences are constant, this confirms that the sequence is cubic. The general form of a cubic sequence is:
\[
a_n = An^3 + Bn^2 + Cn + D
\]

#### Step 4: Use the initial terms to set up equations.
We know the first few terms of the sequence:
\[
\begin{aligned}
&a_1 = 4, \\
&a_2 = 14, \\
&a_3 = 40, \\
&a_4 = 88.
\end{aligned}
\]
Substitute these into the general form:
1. For \( n = 1 \):
\[
A(1)^3 + B(1)^2 + C(1) + D = 4 \implies A + B + C + D = 4
\]
2. For \( n = 2 \):
\[
A(2)^3 + B(2)^2 + C(2) + D = 14 \implies 8A + 4B + 2C + D = 14
\]
3. For \( n = 3 \):
\[
A(3)^3 + B(3)^2 + C(3) + D = 40 \implies 27A + 9B + 3C + D = 40
\]
4. For \( n = 4 \):
\[
A(4)^3 + B(4)^2 + C(4) + D = 88 \implies 64A + 16B + 4C + D = 88
\]

#### Step 5: Solve the system of equations.
We have the following system:
\[
\begin{aligned}
1. & \quad A + B + C + D = 4, \\
2. & \quad 8A + 4B + 2C + D = 14, \\
3. & \quad 27A + 9B + 3C + D = 40, \\
4. & \quad 64A + 16B + 4C + D = 88.
\end{aligned}
\]

Subtract equation 1 from equation 2:
\[
(8A + 4B + 2C + D) - (A + B + C + D) = 14 - 4 \implies 7A + 3B + C = 10 \quad \text{(Equation 5)}
\]

Subtract equation 2 from equation 3:
\[
(27A + 9B + 3C + D) - (8A + 4B + 2C + D) = 40 - 14 \implies 19A + 5B + C = 26 \quad \text{(Equation 6)}
\]

Subtract equation 3 from equation 4:
\[
(64A + 16B + 4C + D) - (27A + 9B + 3C + D) = 88 - 40 \implies 37A + 7B + C = 48 \quad \text{(Equation 7)}
\]

Now we have a new system:
\[
\begin{aligned}
5. & \quad 7A + 3B + C = 10, \\
6. & \quad 19A + 5B + C = 26, \\
7. & \quad 37A + 7B + C = 48.
\end{aligned}
\]

Subtract equation 5 from equation 6:
\[
(19A + 5B + C) - (7A + 3B + C) = 26 - 10 \implies 12A + 2B = 16 \implies 6A + B = 8 \quad \text{(Equation 8)}
\]

Subtract equation 6 from equation 7:
\[
(37A + 7B + C) - (19A + 5B + C) = 48 - 26 \implies 18A + 2B = 22 \implies 9A + B = 11 \quad \text{(Equation 9)}
\]

Subtract equation 8 from equation 9:
\[
(9A + B) - (6A + B) = 11 - 8 \implies 3A = 3 \implies A = 1
\]

Substitute \( A = 1 \) into equation 8:
\[
6(1) + B = 8 \implies 6 + B = 8 \implies B = 2
\]

Substitute \( A = 1 \) and \( B = 2 \) into equation 5:
\[
7(1) + 3(2) + C = 10 \implies 7 + 6 + C = 10 \implies C = -3
\]

Substitute \( A = 1 \), \( B = 2 \), and \( C = -3 \) into equation 1:
\[
1 + 2 - 3 + D = 4 \implies 0 + D = 4 \implies D = 4
\]

Thus, the formula for the \( n \)-th term is:
\[
a_n = n^3 + 2n^2 - 3n + 4
\]

---

Exercise 2:

The sequence is given as:
\[ 4, 23, 66, 145, 272, \ldots \]

#### Step 1: Calculate the first differences.
\[
\begin{aligned}
&23 - 4 = 19, \\
&66 - 23 = 43, \\
&145 - 66 = 79, \\
&272 - 145 = 127.
\end{aligned}
\]
So, the first differences are:
\[ 19, 43, 79, 127, \ldots \]

#### Step 2: Calculate the second differences.
\[
\begin{aligned}
&43 - 19 = 24, \\
&79 - 43 = 36, \\
&127 - 79 = 48.
\end{aligned}
\]
So, the second differences are:
\[ 24, 36, 48, \ldots \]

#### Step 3: Calculate the third differences.
\[
\begin{aligned}
&36 - 24 = 12, \\
&48 - 36 = 12.
\end{aligned}
\]
So, the third differences are:
\[ 12, 12, \ldots \]

Since the third differences are constant, this confirms that the sequence is cubic. The general form is:
\[
a_n = An^3 + Bn^2 + Cn + D
\]

#### Step 4: Use the initial terms to set up equations.
We know the first few terms of the sequence:
\[
\begin{aligned}
&a_1 = 4, \\
&a_2 = 23, \\
&a_3 = 66, \\
&a_4 = 145.
\end{aligned}
\]
Substitute these into the general form:
1. For \( n = 1 \):
\[
A(1)^3 + B(1)^2 + C(1) + D = 4 \implies A + B + C + D = 4
\]
2. For \( n = 2 \):
\[
A(2)^3 + B(2)^2 + C(2) + D = 23 \implies 8A + 4B + 2C + D = 23
\]
3. For \( n = 3 \):
\[
A(3)^3 + B(3)^2 + C(3) + D = 66 \implies 27A + 9B + 3C + D = 66
\]
4. For \( n = 4 \):
\[
A(4)^3 + B(4)^2 + C(4) + D = 145 \implies 64A + 16B + 4C + D = 145
\]

#### Step 5: Solve the system of equations.
We have the following system:
\[
\begin{aligned}
1. & \quad A + B + C + D = 4, \\
2. & \quad 8A + 4B + 2C + D = 23, \\
3. & \quad 27A + 9B + 3C + D = 66, \\
4. & \quad 64A + 16B + 4C + D = 145.
\end{aligned}
\]

Subtract equation 1 from equation 2:
\[
(8A + 4B + 2C + D) - (A + B + C + D) = 23 - 4 \implies 7A + 3B + C = 19 \quad \text{(Equation 5)}
\]

Subtract equation 2 from equation 3:
\[
(27A + 9B + 3C + D) - (8A + 4B + 2C + D) = 66 - 23 \implies 19A + 5B + C = 43 \quad \text{(Equation 6)}
\]

Subtract equation 3 from equation 4:
\[
(64A + 16B + 4C + D) - (27A + 9B + 3C + D) = 145 - 66 \implies 37A + 7B + C = 79 \quad \text{(Equation 7)}
\]

Now we have a new system:
\[
\begin{aligned}
5. & \quad 7A + 3B + C = 19, \\
6. & \quad 19A + 5B + C = 43, \\
7. & \quad 37A + 7B + C = 79.
\end{aligned}
\]

Subtract equation 5 from equation 6:
\[
(19A + 5B + C) - (7A + 3B + C) = 43 - 19 \implies 12A + 2B = 24 \implies 6A + B = 12 \quad \text{(Equation 8)}
\]

Subtract equation 6 from equation 7:
\[
(37A + 7B + C) - (19A + 5B + C) = 79 - 43 \implies 18A + 2B = 36 \implies 9A + B = 18 \quad \text{(Equation 9)}
\]

Subtract equation 8 from equation 9:
\[
(9A + B) - (6A + B) = 18 - 12 \implies 3A = 6 \implies A = 2
\]

Substitute \( A = 2 \) into equation 8:
\[
6(2) + B = 12 \implies 12 + B = 12 \implies B = 0
\]

Substitute \( A = 2 \) and \( B = 0 \) into equation 5:
\[
7(2) + 3(0) + C = 19 \implies 14 + C = 19 \implies C = 5
\]

Substitute \( A = 2 \), \( B = 0 \), and \( C = 5 \) into equation 1:
\[
2 + 0 + 5 + D = 4 \implies 7 + D = 4 \implies D = -3
\]

Thus, the formula for the \( n \)-th term is:
\[
a_n = 2n^3 + 5n - 3
\]

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Final Answers:

\[
\boxed{n^3 + 2n^2 - 3n + 4} \quad \text{and} \quad \boxed{2n^3 + 5n - 3}
\]