Find the ` n^(th)` terms of the series
` 5 + 7 + 13 + 31 + …`

Explanation

To find the \( n^{th} \) term of the series \( 5, 7, 13, 31, \ldots \), we first need to analyze the pattern in the series.

### Step 1: Identify the terms
The given series is:
- \( a_1 = 5 \)
- \( a_2 = 7 \)
- \( a_3 = 13 \)
- \( a_4 = 31 \)

### Step 2: Find the differences between consecutive terms
Let's calculate the differences between consecutive terms:
- \( a_2 - a_1 = 7 - 5 = 2 \)
- \( a_3 - a_2 = 13 - 7 = 6 \)
- \( a_4 - a_3 = 31 - 13 = 18 \)

So, the first differences are:
- \( 2, 6, 18 \)

### Step 3: Find the second differences
Now, let's find the differences of the first differences:
- \( 6 - 2 = 4 \)
- \( 18 - 6 = 12 \)

So, the second differences are:
- \( 4, 12 \)

### Step 4: Find the third differences
Now, let's find the differences of the second differences:
- \( 12 - 4 = 8 \)

The third difference is constant:
- \( 8 \)

### Step 5: Formulate the \( n^{th} \) term
Since the third difference is constant, we can assume that the \( n^{th} \) term can be expressed as a cubic polynomial:
\[ a_n = An^3 + Bn^2 + Cn + D \]

### Step 6: Set up equations using known terms
We can use the first four terms to set up a system of equations:
1. For \( n = 1 \): \( A(1^3) + B(1^2) + C(1) + D = 5 \) → \( A + B + C + D = 5 \)
2. For \( n = 2 \): \( A(2^3) + B(2^2) + C(2) + D = 7 \) → \( 8A + 4B + 2C + D = 7 \)
3. For \( n = 3 \): \( A(3^3) + B(3^2) + C(3) + D = 13 \) → \( 27A + 9B + 3C + D = 13 \)
4. For \( n = 4 \): \( A(4^3) + B(4^2) + C(4) + D = 31 \) → \( 64A + 16B + 4C + D = 31 \)

### Step 7: Solve the system of equations
Now we have the following system of equations:
1. \( A + B + C + D = 5 \) (1)
2. \( 8A + 4B + 2C + D = 7 \) (2)
3. \( 27A + 9B + 3C + D = 13 \) (3)
4. \( 64A + 16B + 4C + D = 31 \) (4)

Subtract (1) from (2):
\[ (8A + 4B + 2C + D) - (A + B + C + D) = 7 - 5 \]
\[ 7A + 3B + C = 2 \quad (5) \]

Subtract (2) from (3):
\[ (27A + 9B + 3C + D) - (8A + 4B + 2C + D) = 13 - 7 \]
\[ 19A + 5B + C = 6 \quad (6) \]

Subtract (3) from (4):
\[ (64A + 16B + 4C + D) - (27A + 9B + 3C + D) = 31 - 13 \]
\[ 37A + 7B + C = 18 \quad (7) \]

Now we have a new system of equations:
1. \( 7A + 3B + C = 2 \) (5)
2. \( 19A + 5B + C = 6 \) (6)
3. \( 37A + 7B + C = 18 \) (7)

### Step 8: Eliminate \( C \)
Subtract (5) from (6):
\[ (19A + 5B + C) - (7A + 3B + C) = 6 - 2 \]
\[ 12A + 2B = 4 \quad (8) \]

Subtract (6) from (7):
\[ (37A + 7B + C) - (19A + 5B + C) = 18 - 6 \]
\[ 18A + 2B = 12 \quad (9) \]

### Step 9: Solve for \( A \) and \( B \)
From (8):
\[ 6A + B = 2 \quad (10) \]

From (9):
\[ 9A + B = 6 \quad (11) \]

Subtract (10) from (11):
\[ (9A + B) - (6A + B) = 6 - 2 \]
\[ 3A = 4 \implies A = \frac{4}{3} \]

Substituting \( A \) back into (10):
\[ 6\left(\frac{4}{3}\right) + B = 2 \]
\[ 8 + B = 2 \implies B = 2 - 8 = -6 \]

### Step 10: Find \( C \) and \( D \)
Substituting \( A \) and \( B \) into (5):
\[ 7\left(\frac{4}{3}\right) + 3(-6) + C = 2 \]
\[ \frac{28}{3} - 18 + C = 2 \]
\[ C = 2 + 18 - \frac{28}{3} = 20 - \frac{28}{3} = \frac{60 - 28}{3} = \frac{32}{3} \]

Now substituting \( A, B, C \) into (1) to find \( D \):
\[ \frac{4}{3} - 6 + \frac{32}{3} + D = 5 \]
\[ \frac{4 - 18 + 32}{3} + D = 5 \]
\[ \frac{18}{3} + D = 5 \implies 6 + D = 5 \implies D = -1 \]

### Final Formula
Thus, the \( n^{th} \) term is:
\[ a_n = \frac{4}{3}n^3 - 6n^2 + \frac{32}{3}n - 1 \]