To find the \( n^{th} \) term of the series \( 1, 3, 8, 16, 27, 41, \ldots \), we can follow these steps:
### Step 1: Identify the series
The given series is:
\[ 1, 3, 8, 16, 27, 41 \]
### Step 2: Find the differences between consecutive terms
Let's calculate the differences between consecutive terms:
- \( 3 - 1 = 2 \)
- \( 8 - 3 = 5 \)
- \( 16 - 8 = 8 \)
- \( 27 - 16 = 11 \)
- \( 41 - 27 = 14 \)
So, the first differences are:
\[ 2, 5, 8, 11, 14 \]
### Step 3: Find the differences of the first differences
Now, let's calculate the differences of the first differences:
- \( 5 - 2 = 3 \)
- \( 8 - 5 = 3 \)
- \( 11 - 8 = 3 \)
- \( 14 - 11 = 3 \)
The second differences are constant:
\[ 3, 3, 3, 3 \]
### Step 4: Identify the type of series
Since the second differences are constant, we can conclude that the original series can be represented by a quadratic polynomial of the form:
\[ T_n = an^2 + bn + c \]
### Step 5: Set up equations based on known terms
We can use the first three terms of the series to set up a system of equations:
1. For \( n = 1 \):
\[ T_1 = a(1)^2 + b(1) + c = 1 \]
\[ a + b + c = 1 \] (Equation 1)
2. For \( n = 2 \):
\[ T_2 = a(2)^2 + b(2) + c = 3 \]
\[ 4a + 2b + c = 3 \] (Equation 2)
3. For \( n = 3 \):
\[ T_3 = a(3)^2 + b(3) + c = 8 \]
\[ 9a + 3b + c = 8 \] (Equation 3)
### Step 6: Solve the system of equations
We will solve these equations step by step.
**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:**
\[
(5a + b) - (3a + b) = 5 - 2
\]
\[
2a = 3 \implies a = \frac{3}{2}
\]
**Substituting \( a \) back into Equation 4:**
\[
3\left(\frac{3}{2}\right) + b = 2
\]
\[
\frac{9}{2} + b = 2 \implies b = 2 - \frac{9}{2} = -\frac{5}{2}
\]
**Substituting \( a \) and \( b \) back into Equation 1:**
\[
\frac{3}{2} - \frac{5}{2} + c = 1
\]
\[
-\frac{2}{2} + c = 1 \implies c = 1 + 1 = 2
\]
### Step 7: Write the general term
Now we have:
- \( a = \frac{3}{2} \)
- \( b = -\frac{5}{2} \)
- \( c = 2 \)
Thus, the \( n^{th} \) term is:
\[
T_n = \frac{3}{2}n^2 - \frac{5}{2}n + 2
\]
### Step 8: Final answer
The final expression for the \( n^{th} \) term of the series is:
\[
T_n = \frac{1}{2}(3n^2 - 5n + 4)
\]
