Find the sum of n-terms:
`[(1/1)+(1^3 +2^3)/2 +(1^3 +2^3 +3^3)/3+....to n -terms`

Answer

To find the sum of the given series up to n terms, we can break down the problem step by step.

### Step 1: Understand the Series
The series is given as:
\[ S_n = \left( \frac{1^3}{1} \right) + \left( \frac{1^3 + 2^3}{2} \right) + \left( \frac{1^3 + 2^3 + 3^3}{3} \right) + \ldots + \left( \frac{1^3 + 2^3 + \ldots + n^3}{n} \right) \]
We can denote the r-th term of the series as:
\[ T_r = \frac{1^3 + 2^3 + \ldots + r^3}{r} \]

### Step 2: Use the Formula for the Sum of Cubes
The sum of the first r cubes is given by the formula:
\[ 1^3 + 2^3 + \ldots + r^3 = \left( \frac{r(r + 1)}{2} \right)^2 \]
Thus, we can rewrite \(T_r\) as:
\[ T_r = \frac{\left( \frac{r(r + 1)}{2} \right)^2}{r} = \frac{r(r + 1)^2}{4} \]

### Step 3: Find the Sum of n Terms
Now, we need to find the sum of \(T_r\) from \(r = 1\) to \(n\):
\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \frac{r(r + 1)^2}{4} \]
This can be simplified as:
\[ S_n = \frac{1}{4} \sum_{r=1}^{n} r(r + 1)^2 \]

### Step 4: Expand and Simplify the Summation
We need to expand \(r(r + 1)^2\):
\[ r(r + 1)^2 = r(r^2 + 2r + 1) = r^3 + 2r^2 + r \]
Thus, we can write:
\[ S_n = \frac{1}{4} \left( \sum_{r=1}^{n} r^3 + 2\sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \right) \]

### Step 5: Use Known Summation Formulas
We will use the following formulas for summations:
1. \(\sum_{r=1}^{n} r = \frac{n(n + 1)}{2}\)
2. \(\sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6}\)
3. \(\sum_{r=1}^{n} r^3 = \left( \frac{n(n + 1)}{2} \right)^2\)

Substituting these into our expression for \(S_n\):
\[ S_n = \frac{1}{4} \left( \left( \frac{n(n + 1)}{2} \right)^2 + 2 \cdot \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \right) \]

### Step 6: Simplify the Expression
Calculating each part:
1. The first term: \(\left( \frac{n(n + 1)}{2} \right)^2 = \frac{n^2(n + 1)^2}{4}\)
2. The second term: \(2 \cdot \frac{n(n + 1)(2n + 1)}{6} = \frac{n(n + 1)(2n + 1)}{3}\)
3. The third term: \(\frac{n(n + 1)}{2}\)

Combining these, we can factor out \(\frac{n(n + 1)}{4}\):
\[ S_n = \frac{n(n + 1)}{4} \left( \frac{n(n + 1)}{4} + \frac{2(2n + 1)}{3} + 2 \right) \]

### Step 7: Final Calculation
Now, we need to combine the terms in the parenthesis and simplify further. This will yield the final expression for \(S_n\).

### Final Answer
After simplifying, we find:
\[ S_n = \frac{n(n + 1)(3n^2 + 11n + 10)}{48} \]