If n ≥ 2 is a positive integer, then the sum of the series ⁿ⁺¹C₂ + 2 ( ²C₂ + ³C₂ + ⁴C₂ + . . . + ⁿC₂

Q: Why can't I just expand every term? It seems safer.

Expanding every term is a trap that leads to a massive, unmanageable summation. In JEE Advanced, the goal is not just to calculate, but to recognize the underlying structure. Combinatorial identities like the Hockey-stick and Pascal's identity are the elegant, professional tools designed to collapse these complex series into simple, beautiful forms. Trust the identity; it is your shortcut to accuracy.

Q: How do I know when to use the Hockey-stick identity versus Pascal's identity?

Think of the geometry of Pascal's Triangle. The Hockey-stick identity is used when you are summing along a diagonal (changing the upper index, constant lower index). Pascal's identity is used when you are combining two adjacent terms in the same row (same upper index, consecutive lower indices). Visualize the triangle, and the path will become clear.

Q: Why did we split the $2({^{n+1}C_3})$ into two separate terms?

We are looking for the structure required by Pascal's Identity, which is ${^nC_r} + {^nC_{r-1}} = {^{n+1}C_r}$. We had ${^{n+1}C_2} + 2({^{n+1}C_3})$. By splitting the coefficient, we created ${^{n+1}C_2} + {^{n+1}C_3} + {^{n+1}C_3}$. This allowed us to pair the first two terms to form ${^{n+2}C_3}$, simplifying the expression significantly.

Question

If $n \ge 2$ is a positive integer, then the sum of the series $^{n+1}C_2 + 2 ( ^2C_2 + ^3C_2 + ^4C_2 + . . . + ^nC_2)$ is :

Options:

(A) $\frac{n(n+1)^2(n+2)}{12}$

(B) $\frac{n(n-1)(2n+1)}{6}$

(C) $\frac{n(n+1)(2n+1)}{6}$

(D) $\frac{n(2n+1)(3n+1)}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion. When you first look at the series

$$S = {^{n+1}C_2} + 2 \left( {^2C_2} + {^3C_2} + \dots + {^nC_2} \right)$$

it is natural to feel a slight hesitation. It looks like a mess of combinations, a sprawling sequence that threatens to consume your time.

But I want you to pause. Take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Our goal is to find that elegance.

### Phase 1: The Hockey-stick Revelation

Look closely at the bracketed term: ${^2C_2} + {^3C_2} + \dots + {^nC_2}$. What do you see? The lower index is fixed at $2$, but the upper index is climbing steadily from $2$ to $n$.

This is the signature of the **Hockey-stick Identity**. Imagine Pascal's Triangle; if you start at the edge and move down a diagonal, summing the terms, the result 'hooks' to the left.

Mathematically, this is expressed as:

$$\sum_{i=r}^{n} {^iC_r} = {^{n+1}C_{r+1}}$$

By applying this to our series, the entire bracket collapses into a single, beautiful term: ${^{n+1}C_3}$. We have simplified the problem significantly, transforming a long summation into:

$$S = {^{n+1}C_2} + 2 \left( {^{n+1}C_3} \right)$$

### Phase 2: The Algebraic Pivot

Now, we face a small hurdle. We have a coefficient of $2$ multiplying our new term, which prevents us from immediately applying Pascal's Identity.

We need to manipulate the expression to fit the identity. We split the term $2({^{n+1}C_3})$ into ${^{n+1}C_3} + {^{n+1}C_3}$.

Now our expression reads:

$$S = {^{n+1}C_2} + {^{n+1}C_3} + {^{n+1}C_3}$$

This is a classic algebraic maneuver. We are setting the stage for the next act.

### Phase 3: The Power of Pascal's Identity

Look at the first two terms: ${^{n+1}C_2} + {^{n+1}C_3}$. They share the same upper index, $n+1$, and their lower indices, $2$ and $3$, are consecutive.

This is the perfect environment for **Pascal's Identity**:

$${^nC_r} + {^nC_{r-1}} = {^{n+1}C_r}$$

Applying this, the sum ${^{n+1}C_2} + {^{n+1}C_3}$ transforms into ${^{n+2}C_3}$. Our series is now:

$$S = {^{n+2}C_3} + {^{n+1}C_3}$$

We have reduced a complex series into the sum of just two combinations. The heavy lifting is done.

### Phase 4: The Final Algebraic Elegance

Now, we move to the final stage: expansion and simplification. We know that:

$${^nC_3} = \frac{n(n-1)(n-2)}{6}$$

Applying this to our terms, we get:

$$S = \frac{(n+2)(n+1)n}{6} + \frac{(n+1)n(n-1)}{6}$$

Do not rush to multiply these out! That is a path to calculation errors. Instead, look for the common factors. Both terms share $\frac{n(n+1)}{6}$.

Let us factor that out:

$$S = \frac{n(n+1)}{6} [ (n+2) + (n-1) ]$$

Inside the bracket, we have $(n+2) + (n-1)$, which simplifies to $2n+1$. Thus, our final result is:

**$$S = \frac{n(n+1)(2n+1)}{6}$$**

This is not just any result; it is the famous formula for the sum of the squares of the first $n$ natural numbers. The symmetry is breathtaking. We started with a series of combinations and ended with a fundamental sum of squares. Keep this perspective, and you will conquer any problem the exam throws at you.

Animated Solution for Mathematics - Binomial Theorem: If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

Select Answer:

Visualized Solution

Analyzing the Given Series $S$

The Hockey-stick Identity

Substituting Back into $S$

Splitting the Term

Applying Pascal's Identity

The Collapsed Series

Expanding the Combinations

Factoring Common Terms

Final Simplification

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n≥2 is a positive integer, then the sum of the series n+1C2​+2(2C2​+3C2​+4C2​+...+nC2​) is :