3 + 13 + 29 + 51 + 79+… to n terms =

Options

A. `2n^(2)+7n^(3)`
B. `n^(2)+5n^(3)`
C. `n^(3)+2n^(2)`
D. none of these

Answer

To find the sum of the series \( S_n = 3 + 13 + 29 + 51 + 79 + \ldots \) up to \( n \) terms, we will first identify a pattern in the terms of the series.

### Step 1: Identify the terms of the series
The given series is:
- \( S_1 = 3 \)
- \( S_2 = 3 + 13 = 16 \)
- \( S_3 = 3 + 13 + 29 = 45 \)
- \( S_4 = 3 + 13 + 29 + 51 = 96 \)
- \( S_5 = 3 + 13 + 29 + 51 + 79 = 175 \)

### Step 2: Find the differences between consecutive terms
Let's denote the terms of the series as \( a_n \):
- \( a_1 = 3 \)
- \( a_2 = 13 \)
- \( a_3 = 29 \)
- \( a_4 = 51 \)
- \( a_5 = 79 \)

Now, let's find the differences:
- \( a_2 - a_1 = 13 - 3 = 10 \)
- \( a_3 - a_2 = 29 - 13 = 16 \)
- \( a_4 - a_3 = 51 - 29 = 22 \)
- \( a_5 - a_4 = 79 - 51 = 28 \)

The first differences are: \( 10, 16, 22, 28 \).

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

The second differences are constant and equal to \( 6 \). This indicates that the original sequence can be expressed as a cubic polynomial.

### Step 4: Assume a polynomial form
Assume:
\[ S_n = an^3 + bn^2 + cn + d \]

### Step 5: Use known values to find coefficients
We know:
- \( S_1 = 3 \)
- \( S_2 = 16 \)
- \( S_3 = 45 \)
- \( S_4 = 96 \)

Setting up the equations:
1. \( a(1)^3 + b(1)^2 + c(1) + d = 3 \) (1)
2. \( a(2)^3 + b(2)^2 + c(2) + d = 16 \) (2)
3. \( a(3)^3 + b(3)^2 + c(3) + d = 45 \) (3)
4. \( a(4)^3 + b(4)^2 + c(4) + d = 96 \) (4)

### Step 6: Solve the system of equations
From these equations, we can solve for \( a, b, c, \) and \( d \).

After solving, we find:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 0 \)
- \( d = 0 \)

Thus, the polynomial becomes:
\[ S_n = n^3 + 2n^2 \]

### Final Result
The sum of the series \( S_n \) up to \( n \) terms is:
\[ S_n = n^3 + 2n^2 \]