The `n^(th)` term of the sequence 4,14,30,52,80,114, . . . ., is

Options

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

Answer

To find the nth term of the sequence 4, 14, 30, 52, 80, 114, ..., we will follow a systematic approach.

### Step 1: Identify the Sequence
The given sequence is:
- \( T_1 = 4 \)
- \( T_2 = 14 \)
- \( T_3 = 30 \)
- \( T_4 = 52 \)
- \( T_5 = 80 \)
- \( T_6 = 114 \)

### Step 2: Find the Differences
Let's calculate the first differences between consecutive terms:
- \( T_2 - T_1 = 14 - 4 = 10 \)
- \( T_3 - T_2 = 30 - 14 = 16 \)
- \( T_4 - T_3 = 52 - 30 = 22 \)
- \( T_5 - T_4 = 80 - 52 = 28 \)
- \( T_6 - T_5 = 114 - 80 = 34 \)

So, the first differences are:
- \( 10, 16, 22, 28, 34 \)

### Step 3: Find the Second Differences
Now, let's calculate the second differences:
- \( 16 - 10 = 6 \)
- \( 22 - 16 = 6 \)
- \( 28 - 22 = 6 \)
- \( 34 - 28 = 6 \)

The second differences are constant and equal to 6, indicating that the original sequence is a quadratic sequence.

### Step 4: Form the Quadratic Equation
Since the nth term of a quadratic sequence can be expressed as:
\[ T_n = an^2 + bn + c \]
We know the second difference is \( 2a = 6 \), thus:
\[ a = 3 \]

### Step 5: Use Known Terms to Find b and c
We can use the first few terms to set up equations:
1. For \( n = 1 \):
\[ T_1 = 3(1)^2 + b(1) + c = 4 \]
\[ 3 + b + c = 4 \quad \text{(Equation 1)} \]

2. For \( n = 2 \):
\[ T_2 = 3(2)^2 + b(2) + c = 14 \]
\[ 12 + 2b + c = 14 \quad \text{(Equation 2)} \]

3. For \( n = 3 \):
\[ T_3 = 3(3)^2 + b(3) + c = 30 \]
\[ 27 + 3b + c = 30 \quad \text{(Equation 3)} \]

### Step 6: Solve the Equations
From Equation 1:
\[ b + c = 1 \quad \text{(1)} \]

From Equation 2:
\[ 2b + c = 2 \quad \text{(2)} \]

Subtract Equation (1) from Equation (2):
\[ (2b + c) - (b + c) = 2 - 1 \]
\[ b = 1 \]

Substituting \( b = 1 \) into Equation (1):
\[ 1 + c = 1 \]
\[ c = 0 \]

### Step 7: Final Expression for Tn
Thus, we have:
\[ T_n = 3n^2 + n + 0 \]
or simply:
\[ T_n = 3n^2 + n \]

### Conclusion
The nth term of the sequence is:
\[ T_n = 3n^2 + n \]