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

Step 1 — Rewrite the general term

Starting from the definition of factorial, $(r+1)! = (r+1)\cdot r!= r\cdot r! + r!$, so

Step 2 — Sum from $r=1$ to $r=n$

Writing out the terms:

$(2!-1!) + (3!-2!) + (4!-3!) + \cdots + \left[(n+1)! - n!\right].$ …

We are asked to sum

Concept and Intuition

Writing each term as a fraction over a power of 10 reveals the structure:

• Term 1: $0.2 = \dfrac{2}{10}$
• Term 2: $0.22 = \dfrac{22}{100}$
• Term 3: $0.022 = \dfrac{22}{1000}$
• Term $k$ ($k \ge 2$): $\dfrac{22}{10^k}$

So from the second term onward the numerator is always 22 while the denominator is a power of 10.

Step-by-step solution

1. Separate the first term.

The bracket has $(n-1)$ terms.

2. Sum the geometric tail.

3. Add the first term.

(using $10^{-(n-1)} = 10 \cdot 10^{-n}$).

4. Match to the options. …

Step 1 — The concept: the sum of a GP.

For a geometric progression with first term $a$ and common ratio $r$ (with $r \ne 1$), the sum to $n$ terms is

Step 2 — Write down both sums.

Replacing $n$ by $2n$ in the same formula (same $a$, same $r$ — it is the same GP, just taken further):

Step 3 — Form the ratio $S_n : S_{2n}$.

The common factor $\dfrac{a}{r-1}$ appears in both numerator and denominator, so it cancels entirely — note that the answer therefore does not depend on $a$ at all, which is a useful check against the options (none of them contains $a$ ✓):

Step 4 — The key algebraic step: factor the denominator as a difference of squares.

Write $r^{2n} = \left(r^{n}\right)^{2}$. Then, using $x^2 - 1 = (x-1)(x+1)$ with $x = r^n$:

This is why the expression simplifies so cleanly — the numerator $\left(r^n - 1\right)$ is one of the two factors of the denominator.

Step 5 — Cancel and finish. …

We are told: in a geometric progression (GP), the sum of the first three terms is 16, and the sum of the next three terms (terms 4, 5, 6) is 128. We need the formula for the sum of the first $n$ terms.

Concept and intuition:

A geometric progression is defined by its first term $a$ and common ratio $r$. The sum of the first $n$ terms is $S_n = a \frac{r^n - 1}{r - 1}$ (for $r \neq 1$).

The trick here: the sum of terms 4–6 is just $r^3$ times the sum of terms 1–3, because each term in the second triplet is the corresponding term in the first triplet multiplied by $r^3$. This gives us $r$ directly. Then we can find $a$ and write the sum formula.

Step-by-step solution:

1. Set up the terms.

   Let the first term be $a$ and the common ratio be $r$. Then:

   • First three terms: $a, ar, ar^2$ Their sum: $a + ar + ar^2 = a(1 + r + r^2) = 16$.
   • Next three terms: $ar^3, ar^4, ar^5$ Their sum: $ar^3 + ar^4 + ar^5 = ar^3(1 + r + r^2) = 128$.

2. Find the common ratio $r$.

   Divide the second sum by the first sum:

The factor $(1 + r + r^2)$ cancels (provided it is nonzero, which it is for positive real $r$), giving:

(We take the real positive cube root; $r = 2$ is the natural choice for a standard GP.)

3. Find the first term $a$. Substitute $r = 2$ into the first sum equation:

4. Write the sum to $n$ terms. …

The numerators $4,10,28$ are $3+1,\,9+1,\,27+1=3^k+1$. So the general term is

Summing $n$ terms:

$S_n=\sum_{k=1}^{n}1+\sum_{k=1}^{n}\left(\frac13\right)^k=n+\frac{\frac13\left(1-3^{-n}\right)}{1-\frac13}=n+\frac12\left(1-\frac1{3^n}\right).$ …

The general term is $\dfrac{k}{(k+1)!}$. Write $k=(k+1)-1$:

So …

Concept: Identify the general term, split by partial fractions, telescope.

The terms are (2²−1)/2, (3²−2)/6, (4²−3)/12, …

Numerator of the nth term: (n+1)² − n. Denominator: 2, 6, 12, … = n(n+1).

T(n) = [(n+1)² − n] / [n(n+1)] = (n² + 2n + 1 − n)/(n(n+1)) = (n² + n + 1)/(n(n+1))

= [n(n+1) + 1]/[n(n+1)] = 1 + 1/[n(n+1)] = 1 + 1/n − 1/(n+1). …

Concept: telescoping using k * k! = (k + 1)! - k!.

Indeed (k + 1)! - k! = k!(k + 1 - 1) = k * k!.

Sum from k = 1 to n:

sum k*k! = sum [(k + 1)! - k!]

= (2! - 1!) + (3! - 2!) + ... + ((n + 1)! - n!)

= (n + 1)! - 1!

= (n + 1)! - 1. …

Concept: each bracket is a geometric series; sum term by term.

The k-th term (k = 1, 2, ..., n) is 1 + 2 + 2^2 + ... + 2^k = 2^(k+1) - 1.

(Check: k = 1 gives 1 + 2 = 3 = 2^2 - 1; k = 2 gives 1 + 2 + 4 = 7 = 2^3 - 1.)

Sum of n terms:

S = sum_{k=1}^{n} (2^(k+1) - 1)

= (2^2 + 2^3 + ... + 2^(n+1)) - n

= [2^2 (2^n - 1)/(2 - 1)] - n …

Numerator $=\sum_{r=1}^{n}r^3=\left[\dfrac{n(n+1)}{2}\right]^2$. Denominator $=$ sum of first $n$ odd numbers $=n^2$.

So the $n$-th term:

Sum of first nine terms: …