Find the `n ^(th)` term of the series `1,3,8,16,27,41,……`

Answer

To find the \( n^{th} \) term of the series \( 1, 3, 8, 16, 27, 41, \ldots \), we will follow these steps:

### Step 1: Identify the differences between the terms
First, we will find the first differences of the series:

- \( 3 - 1 = 2 \)
- \( 8 - 3 = 5 \)
- \( 16 - 8 = 8 \)
- \( 27 - 16 = 11 \)
- \( 41 - 27 = 14 \)

So, the first differences are: \( 2, 5, 8, 11, 14 \).

### Step 2: Find the second differences
Next, we will find the second differences of the first differences:

- \( 5 - 2 = 3 \)
- \( 8 - 5 = 3 \)
- \( 11 - 8 = 3 \)
- \( 14 - 11 = 3 \)

The second differences are constant and equal to \( 3 \).

### Step 3: Formulate the \( n^{th} \) term
Since the second differences are constant, the \( n^{th} \) term can be expressed as a quadratic polynomial:

\[ T_n = an^2 + bn + c \]

### Step 4: Set up equations using known terms
We can use the first three terms of the series to set up equations:

1. For \( n = 1 \): \( T_1 = a(1)^2 + b(1) + c = 1 \) → \( a + b + c = 1 \)
2. For \( n = 2 \): \( T_2 = a(2)^2 + b(2) + c = 3 \) → \( 4a + 2b + c = 3 \)
3. For \( n = 3 \): \( T_3 = a(3)^2 + b(3) + c = 8 \) → \( 9a + 3b + c = 8 \)

### Step 5: Solve the system of equations
We have the following system of equations:

1. \( a + b + c = 1 \) (Equation 1)
2. \( 4a + 2b + c = 3 \) (Equation 2)
3. \( 9a + 3b + c = 8 \) (Equation 3)

Subtract Equation 1 from Equation 2:

\[ (4a + 2b + c) - (a + b + c) = 3 - 1 \]
\[ 3a + b = 2 \quad \text{(Equation 4)} \]

Subtract Equation 2 from Equation 3:

\[ (9a + 3b + c) - (4a + 2b + c) = 8 - 3 \]
\[ 5a + b = 5 \quad \text{(Equation 5)} \]

Now, subtract Equation 4 from Equation 5:

\[ (5a + b) - (3a + b) = 5 - 2 \]
\[ 2a = 3 \quad \Rightarrow \quad a = \frac{3}{2} \]

Substituting \( a = \frac{3}{2} \) into Equation 4:

\[ 3\left(\frac{3}{2}\right) + b = 2 \]
\[ \frac{9}{2} + b = 2 \quad \Rightarrow \quad b = 2 - \frac{9}{2} = -\frac{5}{2} \]

Now substituting \( a \) and \( b \) into Equation 1:

\[ \frac{3}{2} - \frac{5}{2} + c = 1 \]
\[ -\frac{2}{2} + c = 1 \quad \Rightarrow \quad c = 2 \]

### Step 6: Write the final formula for the \( n^{th} \) term
Thus, we have:

\[ T_n = \frac{3}{2}n^2 - \frac{5}{2}n + 2 \]

### Final Answer
The \( n^{th} \) term of the series is:

\[ T_n = \frac{3}{2}n^2 - \frac{5}{2}n + 2 \]