Let `H_(n)=1+(1)/(2)+(1)/(3)+ . . . . .+(1)/(n)`, then the sum to n terms of the series
`(1^(2))/(1^(3))+(1^(2)+2^(2))/(1^(3)+2^(3))+(1^(2)+2^(2)+3^(2))/(1^(3)+2^(3)+3^(3))+ . . . `, is

Options

A. `(4)/(3)H_(n)-1`
B. `(4)/(3)H_(n)+(1)/(n)`
C. `(4)/(3)H_(n)`
D. `(4)/(3)H_(n)+(1)/(n)`

Answer

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

\[ S_n = \frac{1^2}{1^3} + \frac{1^2 + 2^2}{1^3 + 2^3} + \frac{1^2 + 2^2 + 3^2}{1^3 + 2^3 + 3^3} + \ldots + \frac{1^2 + 2^2 + \ldots + n^2}{1^3 + 2^3 + \ldots + n^3} \]

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

\[ T_n = \frac{\sum_{k=1}^{n} k^2}{\sum_{k=1}^{n} k^3} \]

### Step 2: Use the Formulas for Sums
We know that:

- The sum of the squares of the first n natural numbers is given by:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

- The sum of the cubes of the first n natural numbers is given by:

\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 \]

### Step 3: Substitute the Formulas into the General Term
Substituting these formulas into \(T_n\):

\[ T_n = \frac{\frac{n(n+1)(2n+1)}{6}}{\left(\frac{n(n+1)}{2}\right)^2} \]

### Step 4: Simplify the General Term
Now simplify \(T_n\):

\[ T_n = \frac{n(n+1)(2n+1)}{6} \cdot \frac{4}{n^2(n+1)^2} \]

This simplifies to:

\[ T_n = \frac{4(2n+1)}{6n(n+1)} = \frac{2(2n+1)}{3n(n+1)} \]

### Step 5: Find the Sum \(S_n\)
Now we can find the sum \(S_n\):

\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{2(2k+1)}{3k(k+1)} \]

### Step 6: Break Down the Summation
We can separate the summation:

\[ S_n = \frac{2}{3} \sum_{k=1}^{n} \left(\frac{2k+1}{k(k+1)}\right) = \frac{2}{3} \left( \sum_{k=1}^{n} \left(\frac{2}{k} - \frac{2}{k+1}\right) + \sum_{k=1}^{n} \frac{1}{k(k+1)} \right) \]

### Step 7: Evaluate the Summations
The first summation is telescoping:

\[ \sum_{k=1}^{n} \left(\frac{2}{k} - \frac{2}{k+1}\right) = 2(1 - \frac{1}{n+1}) = 2 - \frac{2}{n+1} \]

The second summation can be simplified:

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

### Step 8: Combine the Results
Putting it all together:

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

Simplifying gives:

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

### Final Result
Thus, the sum to n terms of the series is:

\[ S_n = \frac{4}{3} H_n - \frac{2}{3} \]