Find the value of `2/1^3 + 6/(1^3+2^3) + 12/(1^3+2^3+3^3) + 20/(1^3+2^3+3^3+4^3)+.........` upto 60 terms :

Options

A. 2
B. `1/2`
C. 4
D. `1/4`

Answer

To find the value of the series \( S = \frac{2}{1^3} + \frac{6}{1^3 + 2^3} + \frac{12}{1^3 + 2^3 + 3^3} + \frac{20}{1^3 + 2^3 + 3^3 + 4^3} + \ldots \) up to 60 terms, we will first identify the general term of the series and then sum it.

### Step 1: Identify the General Term

The general term \( T_r \) of the series can be expressed as:
\[ T_r = \frac{2r(r+1)}{1^3 + 2^3 + 3^3 + \ldots + r^3} \]

Where:
- The numerator \( 2r(r+1) \) is derived from the pattern observed in the numerators (2, 6, 12, 20, ...).
- The denominator \( 1^3 + 2^3 + \ldots + r^3 \) can be simplified using the formula for the sum of cubes:
\[ 1^3 + 2^3 + \ldots + r^3 = \left(\frac{r(r+1)}{2}\right)^2 \]

### Step 2: Substitute the Denominator

Substituting the formula for the sum of cubes into the general term:
\[ T_r = \frac{2r(r+1)}{\left(\frac{r(r+1)}{2}\right)^2} \]

### Step 3: Simplify the General Term

Now, simplify \( T_r \):
\[ T_r = \frac{2r(r+1)}{\frac{r^2(r+1)^2}{4}} = \frac{2r(r+1) \cdot 4}{r^2(r+1)^2} = \frac{8}{r(r+1)} \]

### Step 4: Write the Series

Now we can write the sum of the first 60 terms:
\[ S_{60} = \sum_{r=1}^{60} T_r = \sum_{r=1}^{60} \frac{8}{r(r+1)} \]

### Step 5: Decompose the Fraction

We can decompose \( \frac{8}{r(r+1)} \) using partial fractions:
\[ \frac{8}{r(r+1)} = \frac{8}{r} - \frac{8}{r+1} \]

### Step 6: Write the Series as a Telescoping Series

Thus, the sum becomes:
\[ S_{60} = 8 \left( \sum_{r=1}^{60} \left( \frac{1}{r} - \frac{1}{r+1} \right) \right) \]

### Step 7: Evaluate the Series

This is a telescoping series:
\[ S_{60} = 8 \left( 1 - \frac{1}{61} \right) = 8 \left( \frac{60}{61} \right) = \frac{480}{61} \]

### Final Result

The value of the series up to 60 terms is:
\[ S_{60} = \frac{480}{61} \]