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)-(2)/(3)`

Answer

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

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

### Step 1: Identify the nth term of the series

The nth term \( T_n \) can be expressed as:

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

### Step 2: Use the formulas for the sums of squares and cubes

We know the formulas for the sums of squares and cubes:

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

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

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

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

### Step 3: Substitute the formulas into the nth 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 nth term

Now, simplify \( T_n \):

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

This simplifies to:

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

Cancelling \( n(n+1) \):

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

### Step 5: Find the sum \( S_n \)

Now, we need to find the sum \( S_n = \sum_{k=1}^{n} T_k \):

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

### Step 6: Split the summation

We can split the fraction:

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

### Step 7: Evaluate the telescoping series

This is a telescoping series:

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

The second sum can be rewritten:

\[ S_n = \frac{2}{3} \left( 2H_n - 2(H_n - 1) \right) = \frac{2}{3} \left( 2H_n - 2H_n + 2 \right) = \frac{2}{3} \cdot 2 = \frac{4}{3} \]

### Final Answer

Thus, the sum to n terms of the series is:

\[ \boxed{\frac{4}{3}} \]