`n^(th)` term of the series `4+14+30+52+ .......=`

Options

A. 5n - 1
B. `2n^(2) + 2n`
C. `3n^(2)+n`
D. `2n^(2)+2`

Answer

To find the \( n^{th} \) term of the series \( 4, 14, 30, 52, \ldots \), we will follow these steps:

### Step 1: Identify the sequence
The given series is:
- \( a_1 = 4 \)
- \( a_2 = 14 \)
- \( a_3 = 30 \)
- \( a_4 = 52 \)

### Step 2: Calculate the differences between consecutive terms
Let's find the first differences:
- \( a_2 - a_1 = 14 - 4 = 10 \)
- \( a_3 - a_2 = 30 - 14 = 16 \)
- \( a_4 - a_3 = 52 - 30 = 22 \)

Now, let's calculate the second differences:
- \( 16 - 10 = 6 \)
- \( 22 - 16 = 6 \)

Since the second differences are constant, the sequence is quadratic.

### Step 3: Assume a quadratic form for the \( n^{th} \) term
We can assume the \( n^{th} \) term \( T_n \) can be expressed as:
\[ T_n = an^2 + bn + c \]
where \( a, b, c \) are constants to be determined.

### Step 4: Set up equations using known terms
Using the first three terms of the series:
1. For \( n = 1 \):
\[ T_1 = a(1)^2 + b(1) + c = 4 \quad \text{(1)} \]
2. For \( n = 2 \):
\[ T_2 = a(2)^2 + b(2) + c = 14 \quad \text{(2)} \]
3. For \( n = 3 \):
\[ T_3 = a(3)^2 + b(3) + c = 30 \quad \text{(3)} \]

### Step 5: Solve the system of equations
From equation (1):
\[ a + b + c = 4 \]
From equation (2):
\[ 4a + 2b + c = 14 \]
From equation (3):
\[ 9a + 3b + c = 30 \]

Now, we can subtract equation (1) from equation (2):
\[ (4a + 2b + c) - (a + b + c) = 14 - 4 \]
This simplifies to:
\[ 3a + b = 10 \quad \text{(4)} \]

Next, subtract equation (2) from equation (3):
\[ (9a + 3b + c) - (4a + 2b + c) = 30 - 14 \]
This simplifies to:
\[ 5a + b = 16 \quad \text{(5)} \]

Now, we can subtract equation (4) from equation (5):
\[ (5a + b) - (3a + b) = 16 - 10 \]
This simplifies to:
\[ 2a = 6 \implies a = 3 \]

### Step 6: Substitute \( a \) back to find \( b \) and \( c \)
Substituting \( a = 3 \) into equation (4):
\[ 3(3) + b = 10 \implies 9 + b = 10 \implies b = 1 \]

Now substituting \( a \) and \( b \) into equation (1):
\[ 3 + 1 + c = 4 \implies c = 0 \]

### Step 7: Write the final expression for \( T_n \)
Thus, the \( n^{th} \) term is:
\[ T_n = 3n^2 + n \]

### Conclusion
The \( n^{th} \) term of the series \( 4 + 14 + 30 + 52 + \ldots \) is given by:
\[ T_n = 3n^2 + n \]