For 2 ≤ r ≤ n, ⁿCᵣ + 2ⁿCᵣ₋₁ + ⁿCᵣ₋₂ = (A) ⁿ⁺¹Cᵣ₋₁, (B) 2ⁿ⁺¹Cᵣ₊₁, (C) 2ⁿ⁺²Cᵣ, (D) ⁿ⁺²Cᵣ

Q: Wait, why do we reject the factorial expansion method? Can't we just write out the factorials and simplify?

You absolutely could! Expanding $\binom{n}{r}$ as $\frac{n!}{r!(n-r)!}$ is mathematically valid, but it is a trap in competitive exams like JEE. It leads to messy algebraic manipulation where you have to find common denominators and factor out terms. Using Pascal's Identity is the 'elegant' path—it's faster, less prone to calculation errors, and demonstrates a deeper understanding of the combinatorial structure.

Q: How did you know to split the middle term $2\binom{n}{r-1}$? Is there a rule for that?

That is a fantastic question! In combinatorics, when you see a coefficient like $2$ in a sum of binomial coefficients, it is almost always a signal to split it. Think of it as 'reverse engineering' the identity. We know Pascal's Identity works for coefficients of $1$. By splitting the $2$, we create the necessary structure to apply the identity twice. It is a classic technique in JEE problems to manipulate the expression to fit a known theorem.

Q: Is there a general formula for this? Like, does $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$ always equal $\binom{n+2}{r}$?

Yes, exactly! You have just derived a specific case of the identity $\binom{n}{k} + 2\binom{n}{k-1} + \binom{n}{k-2} = \binom{n+2}{k}$. It is a beautiful property that shows how binomial coefficients grow. You can think of it as moving two steps down in Pascal's triangle. Keep this result in your toolkit—it is a powerful shortcut for similar problems!

Question

For $2 \le r \le n$, $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2} =$

Options:

(A) $\binom{n+1}{r-1}$

(B) $2\binom{n+1}{r+1}$

(C) $2\binom{n+2}{r}$

(D) $\binom{n+2}{r}$

Sigma Percentile · Accepted Answer

### The Elegance of Pascal's Triangle

Welcome, future engineers! Today, we are going to embark on a journey through the heart of combinatorics. We are looking at the expression $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$.

At first glance, it might look like a random collection of terms, but I want you to see the hidden symmetry here. This problem is not just about algebra; it is about the beautiful, recursive nature of Pascal's Triangle.

### The Power of Pascal's Identity

To solve this, we need to summon our most trusted tool: **Pascal's Identity**. It is the fundamental law that governs how the triangle is built:

$$
\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}
$$

This identity is profound. It tells us that if we take two adjacent numbers in a row of Pascal's triangle, their sum is the number directly below them in the next row.

The upper index $n$ increments by $1$, and we take the larger of the two lower indices. This is the geometric reality we are working with.

### The 'Aha!' Moment: Splitting the Middle

Now, look at our expression: $\binom{n}{r} + 2\binom{n}{r-1} + \binom{n}{r-2}$. The middle term, $2\binom{n}{r-1}$, is the obstacle.

Pascal's Identity requires coefficients of $1$. How do we break this deadlock? We use a simple, yet brilliant algebraic maneuver: we split the middle term.

Just as we might write $2x$ as $x + x$, we write:

$$
2\binom{n}{r-1} = \binom{n}{r-1} + \binom{n}{r-1}
$$

By doing this, our expression transforms into:

$$
\binom{n}{r} + \binom{n}{r-1} + \binom{n}{r-1} + \binom{n}{r-2}
$$

Suddenly, the path forward becomes clear. We have created two pairs that are perfectly suited for Pascal's Identity.

### The Final Collapse

Let's group these terms:

$$
\left(\binom{n}{r} + \binom{n}{r-1}\right) + \left(\binom{n}{r-1} + \binom{n}{r-2}\right)
$$

Applying the identity to the first bracket, we get $\binom{n+1}{r}$. Applying it to the second bracket, we get $\binom{n+1}{r-1}$.

Our expression has now collapsed into:

$$
\binom{n+1}{r} + \binom{n+1}{r-1}
$$

Look at that! We have the exact same structure we started with, but now with $n+1$ instead of $n$. We apply Pascal's Identity one final time.

The upper index increases to $n+2$, and we keep the larger lower index, $r$. The result is:

$$
\binom{n+2}{r}
$$

And there it is—the elegance of the final answer. We started with four terms and, through the recursive beauty of Pascal's Identity, we arrived at a single, clean binomial coefficient.

This is the kind of mathematical harmony that makes physics and engineering so rewarding. Keep practicing this, and you will start to see these patterns everywhere!

Animated Solution for Mathematics - Binomial Theorem: For $2 \leq r \leq n$ , $(r n) + 2 (r - 1 n) + (r - 2 n) =$

Select Answer:

Visualized Solution

The Given Expression

Pascal's Identity

Analyzing the Middle Term

Splitting $2 (r - 1 n)$

The Expanded Expression

Grouping the Terms

Simplifying the First Group

Simplifying the Second Group

Combining the Results

Final Application of Pascal's Identity

The Final Answer

The Sigma Insight: Properties of Binomial Coefficients

The Elegance of Pascal's Triangle

The Power of Pascal's Identity

The 'Aha!' Moment

The Final Collapse

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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; } For 2≤r≤n, (rn​)+2(r−1n​)+(r−2n​)=