Let be a sequence such that a1 + a2 + ... + an = (n² + 3n) / ((n+1)(n+2)). If 28 Σ(k=1 to 10) 1

Q: Wait, can't I just calculate $a_1, a_2, a_3$ and guess the pattern instead of doing all this algebra?

While intuition is a superpower, in JEE Advanced, we need rigor. Calculating the first few terms is a fantastic way to verify your work or gain a 'feel' for the sequence, but deriving the general term $a_n$ ensures you don't miss a subtle shift in the pattern that might occur at higher values of $n$. Trust the algebra; it is the safest path to the truth.

Q: The algebra for $S_n - S_{n-1}$ looks terrifying. Did I make a mistake?

It looks daunting, but trust the process. The terms are structured to cancel out. If you find yourself with a massive polynomial, take a breath, re-check your common denominator, and look for the cancellation of the $n^3$ and $n^2$ terms. These problems are designed to be elegant, not messy. If it feels like a mess, you are likely one step away from a beautiful simplification.

Q: Why does the problem ask for $p_1 p_2 \dots p_m$? Is there a trick?

It is a beautiful way to test your prime factorization skills. Once you have the final number, the prime factors will naturally emerge. If you don't get a sequence of consecutive primes, re-check your arithmetic in the summation step! The problem is designed to reward you with a clean, sequential product of primes if your logic is sound.

Question

Let $\langle a_n \rangle$ be a sequence such that $a_1 + a_2 + \dots + a_n = \frac{n^2 + 3n}{(n+1)(n+2)}$. If $28 \sum_{k=1}^{10} \frac{1}{a_k} = p_1 p_2 p_3 \dots p_m$, where $p_1, p_2, \dots, p_m$ are the first $m$ prime numbers, then $m$ is equal to

Options:

(A) 5

(B) 8

(C) 6

(D) 7

Sigma Percentile · Accepted Answer

### The Architecture of a Sequence: A Journey into Elegance

Welcome, student. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure. When you look at a sequence problem in the JEE Advanced, it is easy to feel overwhelmed by the symbols.

Think of this problem as a puzzle box. The formula for the sum $S_n$ is the box, and our goal is to find the hidden mechanism—the $n$-th term $a_n$—that makes it tick.

### Phase 1: The Detective Work

We are given the sum of the first $n$ terms:

$$S_n = \frac{n^2 + 3n}{(n+1)(n+2)}$$

This is our starting point. It is a rational expression, and it holds the entire history of the sequence within it. But we don't want the history; we want the present.

We want the $n$-th term, $a_n$. How do we isolate it? We use the fundamental relation:

$$a_n = S_n - S_{n-1}$$

Imagine you have a stack of books. $S_n$ is the total height of $n$ books. $S_{n-1}$ is the height of the first $n-1$ books. If you take the total height and subtract the height of the first $n-1$ books, what remains? Exactly, the $n$-th book!

### Phase 2: The Algebraic Crucible

Now, we enter the crucible. We must substitute the expressions. For $S_n$, we use the given formula. For $S_{n-1}$, we replace every $n$ with $(n-1)$.

The expression for $S_{n-1}$ becomes:

$$S_{n-1} = \frac{(n-1)^2 + 3(n-1)}{((n-1)+1)((n-1)+2)} = \frac{n^2 - 2n + 1 + 3n - 3}{n(n+1)} = \frac{n^2 + n - 2}{n(n+1)}$$

Now, we subtract:

$$a_n = \frac{n^2 + 3n}{(n+1)(n+2)} - \frac{n^2 + n - 2}{n(n+1)}$$

To perform this subtraction, we need a common denominator, which is $n(n+1)(n+2)$. When we expand the numerators and subtract, the complexity collapses into:

$$a_n = \frac{4}{n(n+1)(n+2)}$$

### Phase 3: The Summation Mastery

With $a_n$ in hand, the problem asks for the sum of the reciprocals:

$$\sum_{k=1}^{10} \frac{1}{a_k}$$

Since $a_k = \frac{4}{k(k+1)(k+2)}$, the reciprocal is simply:

$$\frac{1}{a_k} = \frac{k(k+1)(k+2)}{4}$$

We need to calculate:

$$\frac{1}{4} \sum_{k=1}^{10} k(k+1)(k+2)$$

Here, we invoke a powerful theorem for the sum of products of consecutive integers. The sum of the product of $r$ consecutive integers is given by:

$$\sum_{k=1}^n k(k+1)\dots(k+r-1) = \frac{n(n+1)\dots(n+r)}{r+1}$$

In our case, we have three consecutive integers ($r=3$), so the sum is:

$$\frac{n(n+1)(n+2)(n+3)}{4}$$

Substituting $n=10$, we get:

$$\frac{10 \cdot 11 \cdot 12 \cdot 13}{4}$$

Don't forget the factor of $\frac{1}{4}$ we pulled out earlier! Our total sum is:

$$\frac{1}{4} \cdot \frac{10 \cdot 11 \cdot 12 \cdot 13}{4} = \frac{10 \cdot 11 \cdot 3 \cdot 13}{4} = \frac{5 \cdot 11 \cdot 3 \cdot 13}{2}$$

### Phase 4: The Prime Finale

The problem asks for $28$ times this sum. Let us multiply:

$$28 \cdot \frac{5 \cdot 11 \cdot 3 \cdot 13}{2} = 14 \cdot 5 \cdot 11 \cdot 3 \cdot 13$$

We need to express this as a product of the first $m$ prime numbers. Let us break down $14$ into its prime components: $14 = 2 \cdot 7$.

Now, our product is:

$$(2 \cdot 7) \cdot 5 \cdot 11 \cdot 3 \cdot 13$$

Arranging them in ascending order:

$$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$$

These are the first **6** prime numbers! Thus, **$m=6$**.

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Visualized Solution

Given Sum $S_{n}$

Finding the $n$ -th Term

Substitution

Simplifying $S_{n - 1}$

Common Denominator

Simplifying the Numerator

Reciprocal of $a_{k}$

Setting up the Summation

Summation Formula

Evaluating for $n = 10$

Multiplying by $28$

Prime Factorization

Final Value of $m$

The Sigma Insight: Sum of Special Series

The Architecture of a Sequence

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Given Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-th Term

Substitution

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn−1​

Common Denominator

Simplifying the Numerator

Reciprocal of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ak​

Setting up the Summation

Summation Formula

Evaluating for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=10

Multiplying by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 28