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 infinite terms

Options

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

Answer

To find the value of the series

\[ S = \frac{2}{1^3} + \frac{6}{1^3 + 2^3} + \frac{12}{1^3 + 2^3 + 3^3} + \frac{20}{1^3 + 2^3 + 3^3 + 4^3} + \ldots \]

we will analyze the general term and then sum it up to infinity.

### Step 1: Identify the General Term

The first term can be expressed as:

\[ T_1 = \frac{2}{1^3} = \frac{1 \cdot 2}{1^3} \]

The second term can be expressed as:

\[ T_2 = \frac{6}{1^3 + 2^3} = \frac{2 \cdot 3}{1^3 + 2^3} \]

The third term can be expressed as:

\[ T_3 = \frac{12}{1^3 + 2^3 + 3^3} = \frac{3 \cdot 4}{1^3 + 2^3 + 3^3} \]

The fourth term can be expressed as:

\[ T_4 = \frac{20}{1^3 + 2^3 + 3^3 + 4^3} = \frac{4 \cdot 5}{1^3 + 2^3 + 3^3 + 4^3} \]

From this pattern, we can see that the general term \( T_r \) can be written as:

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

### Step 2: Sum of Cubes Formula

We know that the sum of the first \( r \) cubes is given by:

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

### Step 3: Substitute the Sum of Cubes

Now substituting this into our general term \( T_r \):

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

This simplifies to:

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

### Step 4: Rewrite the Series

Thus, the series \( S \) can be rewritten as:

\[ S = \sum_{r=1}^{\infty} T_r = \sum_{r=1}^{\infty} \frac{4}{r(r + 1)} \]

### Step 5: Simplify the Series

We can simplify \( \frac{4}{r(r + 1)} \) using partial fractions:

\[ \frac{4}{r(r + 1)} = 4 \left( \frac{1}{r} - \frac{1}{r + 1} \right) \]

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

Now, substituting this back into the series:

\[ S = 4 \sum_{r=1}^{\infty} \left( \frac{1}{r} - \frac{1}{r + 1} \right) \]

This is a telescoping series, and when we write out the first few terms, we see:

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

### Step 7: Evaluate the Limit

In the limit as \( n \) approaches infinity, all intermediate terms cancel out, leaving:

\[ S = 4 \left( 1 - 0 \right) = 4 \]

### Final Answer

Thus, the value of the series is:

\[ \boxed{4} \]