The sum of `(1/2 . 2/2)/(1^3) + (2/2 . 3/2)/(1^3 + 2^3) + (3/2 . 4/2)/(1^3 + 2^3 + 3^3) + `….. Upto n terms is equal to

Options

A. `(n-1)/(n)`
B. `(n)/(n+1)`
C. `(n+1)/(n+2)`
D. `(n+1)/(n)`

Answer

To solve the given problem, we need to find the sum of the series:

\[ S = \frac{(1/2) \cdot (2/2)}{1^3} + \frac{(2/2) \cdot (3/2)}{1^3 + 2^3} + \frac{(3/2) \cdot (4/2)}{1^3 + 2^3 + 3^3} + \ldots \text{ up to } n \text{ terms} \]

### Step 1: Identify the General Term
The general term of the series can be expressed as:

\[ T_r = \frac{(r/2) \cdot ((r+1)/2)}{1^3 + 2^3 + \ldots + r^3} \]

### Step 2: Simplify the General Term
The denominator \(1^3 + 2^3 + \ldots + r^3\) can be simplified using the formula for the sum of cubes:

\[ 1^3 + 2^3 + \ldots + r^3 = \left(\frac{r(r+1)}{2}\right)^2 \]

Thus, we can rewrite \(T_r\):

\[ T_r = \frac{(r/2) \cdot ((r+1)/2)}{\left(\frac{r(r+1)}{2}\right)^2} \]

### Step 3: Further Simplification
Now, substituting the sum of cubes into the general term:

\[ T_r = \frac{(r(r+1)/4)}{\left(\frac{r(r+1)}{2}\right)^2} = \frac{(r(r+1)/4)}{\frac{r^2(r+1)^2}{4}} = \frac{1}{r(r+1)} \]

### Step 4: Sum the Series
Now we need to sum \(T_r\) from \(r=1\) to \(n\):

\[ S = \sum_{r=1}^{n} \frac{1}{r(r+1)} \]

### Step 5: Use Partial Fraction Decomposition
We can express \(\frac{1}{r(r+1)}\) using partial fractions:

\[ \frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1} \]

### Step 6: Write the Sum
Now substituting this back into the sum:

\[ S = \sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) \]

### Step 7: Evaluate the Sum
This is a telescoping series. Most terms will cancel out:

\[ S = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \]

After cancellation, we are left with:

\[ S = 1 - \frac{1}{n+1} \]

### Step 8: Final Result
Thus, the sum \(S\) can be simplified to:

\[ S = \frac{n}{n+1} \]

### Conclusion
The final answer for the sum of the series is:

\[ \boxed{\frac{n}{n+1}} \]