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 can follow these steps:

### Step 1: Identify the General Term
The series can be expressed in terms of its general term. The first term is \(1^3\), the second term is \(-(1.5)^3\), the third term is \(2^3\), and the fourth term is \(-(2.5)^3\).

We can observe that the general term \(T_r\) can be defined as:
\[ T_r = (-1)^{r+1} \left( \left( \frac{r}{2} \right)^3 + \left( \frac{r}{2} + 0.5 \right)^3 \right) \]
However, we can simplify this to:
\[ T_r = r^3 - (r + 0.5)^3 \]

### Step 2: Simplify the General Term
Now, let's simplify \(T_r\):
\[ T_r = r^3 - (r + 0.5)^3 \]
Expanding \((r + 0.5)^3\):
\[ (r + 0.5)^3 = r^3 + 3 \cdot r^2 \cdot 0.5 + 3 \cdot r \cdot (0.5)^2 + (0.5)^3 = r^3 + \frac{3}{2}r^2 + \frac{3}{4}r + \frac{1}{8} \]
Thus,
\[ T_r = r^3 - \left( r^3 + \frac{3}{2}r^2 + \frac{3}{4}r + \frac{1}{8} \right) \]
\[ T_r = -\frac{3}{2}r^2 - \frac{3}{4}r - \frac{1}{8} \]

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

Using the formulas for the sums:
1. \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\)
2. \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\)
3. \(\sum_{r=1}^{n} 1 = n\)

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

### Step 4: Combine and Simplify
Now, let's simplify:
\[ S_n = -\frac{3n(n+1)(2n+1)}{12} - \frac{3n(n+1)}{8} - \frac{n}{8} \]
Finding a common denominator (which is 24):
\[ S_n = -\frac{6n(n+1)(2n+1)}{24} - \frac{9n(n+1)}{24} - \frac{3n}{24} \]
\[ S_n = -\frac{6n(n+1)(2n+1) + 9n(n+1) + 3n}{24} \]

### Final Expression
This gives us the final expression for the sum of the first n terms:
\[ S_n = -\frac{6n(n+1)(2n+1) + 9n(n+1) + 3n}{24} \]