Q.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 $\displaystyle\sum_{r=1}^{n} \dfrac{S_r}{s_r}$ equals
(A) $\dfrac{n(n+1)(n+2)}{6}$
(B) $\dfrac{n(n+1)}{2}$
(C) $\dfrac{n^2+3n+2}{2}$
(D) None of these

The problem involves simplifying a ratio of sums of powers and then summing the resulting expression. The key idea is to use the standard formulas for the sum of the first $n$ natural numbers and the sum of the cubes of the first $n$ natural numbers.

First, recall the formulas:

$S_n = \sum_{k=1}^{n} k^3 = \left(\dfrac{n(n+1)}{2}\right)^2$

$s_n = \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$

Now, simplify the ratio $\dfrac{S_r}{s_r}$:

$\dfrac{S_r}{s_r} = \dfrac{\left(\frac{r(r+1)}{2}\right)^2}{\frac{r(r+1)}{2}} = \dfrac{r(r+1)}{2}$

Finally, sum this expression from $r=1$ to $n$:

$\displaystyle\sum_{r=1}^{n} \dfrac{S_r}{s_r} = \sum_{r=1}^{n} \dfrac{r(r+1)}{2} = \dfrac{1}{2} \sum_{r=1}^{n} (r^2+r)$

$= \dfrac{1}{2} \left( \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \right)$

Using the formulas $\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^{n} r = \dfrac{n(n+1)}{2}$: …

The core of this problem lies in recognizing and correctly applying the standard summation formulas for natural numbers and their cubes. Once these are in place, the task simplifies to algebraic manipulation and evaluating another standard sum.

1. Recall the definitions and standard formulas for $S_n$ and $s_n$.

   The problem defines $S_n$ as the sum of the cubes of the first $n$ natural numbers and $s_n$ as the sum of the first $n$ natural numbers. These are fundamental results in sequences and series.

   The sum of the first $n$ natural numbers:

   $s_n = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

   The sum of the cubes of the first $n$ natural numbers:

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

2. Simplify the ratio $\dfrac{S_r}{s_r}$.

   The problem asks us to sum $\dfrac{S_r}{s_r}$. Before we can sum it, we need to find a simpler expression for this ratio. We substitute $n$ with $r$ in the formulas from Step 1.

Notice that the numerator is simply the square of the denominator. This allows for a direct simplification:

3. Substitute the simplified ratio into the main summation. Now that we have a simpler expression for $\dfrac{S_r}{s_r}$, we can substitute it back into the summation we need to evaluate:

We can factor out the constant $\frac{1}{2}$ from the summation:

4. Evaluate the final summation. We can split the summation into two parts:

Now, we need to recall the formulas for the sum of the first $n$ natural numbers (which is $s_n$ again) and the sum of the squares of the first $n$ natural numbers.

> [!FORMULA]
> The sum of the first $n$ natural numbers:
> $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ …