Let `S_(n)` denote the sum of the cubes of the first n natural numbers and `s_(n)` denote the sum of first n natural numbers then `(S_(n))/(s_(n))` is equal to
(i) `(n(n+1)(n+2))/6`
(ii) `(n( n+1))/2`
(iii) `(n^(2)+3n+2)/2`
(iv) `(2n+1)/3`

Options

A. `(n(n+1)(n+2))/6`
B. `(n( n+1))/2`
C. `(n^(2)+3n+2)/2`
D. `(2n+1)/3`

Answer

To solve the problem, we need to find the ratio \( \frac{S_n}{s_n} \), 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 formula for the sum of the cubes of the first \( n \) natural numbers is:

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

The formula for the sum of the first \( n \) natural numbers is:

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

### Step 2: Substitute the formulas into the ratio

Now we substitute these formulas into the ratio \( \frac{S_n}{s_n} \):

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

### Step 3: Simplify the expression

We can simplify this expression:

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

### Step 4: Final expression

Thus, we have:

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

### Conclusion

The answer is option (ii) \( \frac{n(n+1)}{2} \).

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