let `S_(n)` denote the sum of the cubes of the first n natural numbers and `s_(n)` denote the sum of the first n natural numbers , then `sum_(r=1)^(n)(S_(r))/(s_(r))` equals to

Options

A. `(n(n+1)(n+2))/(6)`
B. `(n(n+1))/(2)`
C. `(n^(2)+3n+2)/(2)`
D. None of these

Explanation

To solve the problem, we need to find the value of the expression:

\[ \sum_{r=1}^{n} \frac{S_r}{s_r} \]

where \( S_n \) is the sum of the cubes of the first \( n \) natural numbers and \( s_n \) is the sum of the first \( n \) natural numbers.

### Step 1: Write the formulas for \( S_n \) and \( s_n \)

The sum of the first \( n \) natural numbers is given by:

\[ s_n = \frac{n(n + 1)}{2} \]

The sum of the cubes of the first \( n \) natural numbers is given by:

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

### Step 2: Substitute \( S_r \) and \( s_r \) into the expression

Now we can substitute these formulas into our expression:

\[ \sum_{r=1}^{n} \frac{S_r}{s_r} = \sum_{r=1}^{n} \frac{\left( \frac{r(r + 1)}{2} \right)^2}{\frac{r(r + 1)}{2}} = \sum_{r=1}^{n} \frac{r(r + 1)}{2} \]

### Step 3: Simplify the expression

The expression simplifies to:

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

### Step 4: Expand \( r(r + 1) \)

We can expand \( r(r + 1) \):

\[ r(r + 1) = r^2 + r \]

Thus, we have:

\[ \sum_{r=1}^{n} r(r + 1) = \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

### Step 5: Use known formulas for \( \sum r^2 \) and \( \sum r \)

We know the formulas for these sums:

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

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

### Step 6: Substitute these formulas back into the expression

Now substituting these back, we have:

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

### Step 7: Combine the terms

To combine these, we can find a common denominator:

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

### Step 8: Factor out the expression

This can be factored as:

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

### Final Result

Thus, the final result for the original expression is:

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