Sum to n terms of the series `1^(3) - (1.5)^(3) +2^(3)-(2.5)^(3) +….` is

Options

A. `1/6 (n+1)^(2)(n+2)^(2)-1/4`
B. `1/32 (n+1)^(2) (n+2)^(2) -1/8`
C. `(n+1)^(2)(n+2)^(2)-35`
D. none of these

Answer

To find the sum to n terms of the series \( 1^3 - (1.5)^3 + 2^3 - (2.5)^3 + \ldots \), we will first identify the general term of the series and then compute the sum.

### Step 1: Identify the General Term
The series can be expressed in terms of its general term \( T_r \):
\[ T_r = r^3 - (0.5 + r)^3 \]
Where \( r \) takes values from 1 to \( n \).

### Step 2: Expand the General Term
Now, we will expand \( (0.5 + r)^3 \):
\[ (0.5 + r)^3 = (0.5)^3 + 3(0.5)^2 r + 3(0.5) r^2 + r^3 \]
Calculating \( (0.5)^3 \):
\[ (0.5)^3 = \frac{1}{8} \]
Calculating \( 3(0.5)^2 = 3 \times \frac{1}{4} = \frac{3}{4} \):
\[ 3(0.5)^2 r = \frac{3}{4}r \]
Calculating \( 3(0.5) = \frac{3}{2} \):
\[ 3(0.5) r^2 = \frac{3}{2} r^2 \]
Putting it all together:
\[ (0.5 + r)^3 = \frac{1}{8} + \frac{3}{4}r + \frac{3}{2}r^2 + r^3 \]

### Step 3: Substitute Back into the General Term
Now substituting back into \( T_r \):
\[ T_r = r^3 - \left(\frac{1}{8} + \frac{3}{4}r + \frac{3}{2}r^2 + r^3\right) \]
This simplifies to:
\[ T_r = -\frac{1}{8} - \frac{3}{4}r - \frac{3}{2}r^2 \]

### Step 4: Sum the General Term
Now we need to sum \( T_r \) from \( r = 1 \) to \( n \):
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \left(-\frac{1}{8} - \frac{3}{4}r - \frac{3}{2}r^2\right) \]
This can be separated into three sums:
\[ S_n = -\frac{1}{8} \sum_{r=1}^{n} 1 - \frac{3}{4} \sum_{r=1}^{n} r - \frac{3}{2} \sum_{r=1}^{n} r^2 \]

### Step 5: Use the Formulas for Summation
Using the formulas:
- \( \sum_{r=1}^{n} 1 = n \)
- \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)
- \( \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \)

Substituting these into the equation:
\[ S_n = -\frac{1}{8} n - \frac{3}{4} \cdot \frac{n(n+1)}{2} - \frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} \]

### Step 6: Simplify the Expression
Now we will simplify each term:
1. The first term is:
\[ -\frac{1}{8} n \]
2. The second term is:
\[ -\frac{3}{4} \cdot \frac{n(n+1)}{2} = -\frac{3}{8} n(n+1) \]
3. The third term is:
\[ -\frac{3}{2} \cdot \frac{n(n+1)(2n+1)}{6} = -\frac{n(n+1)(2n+1)}{4} \]

Combining these:
\[ S_n = -\frac{1}{8} n - \frac{3}{8} n(n+1) - \frac{n(n+1)(2n+1)}{4} \]

### Final Step: Combine and Factor
Combine all these terms to get the final expression for \( S_n \).