The sum of the coefficients of x⁴⁹⁹ and x⁵⁰⁰ in (1 + x)¹⁰⁰⁰ + x(1 + x)⁹⁹⁹ + x²(1 + x)⁹⁹⁸ + ... + x¹⁰

Q: Wait, why is the number of terms $n=1001$ and not $1000$?

That is a fantastic question! It is the most common trap in this problem. Look at the powers of $x$: they start at $x^0$ (the first term) and go up to $x^{1000}$ (the last term). If you count the integers from $0$ to $1000$, you have $1000 - 0 + 1 = 1001$ terms. Always count the indices carefully!

Q: Why can we just ignore the $x^{1001}$ term when finding the coefficients?

We are looking for the coefficients of $x^{499}$ and $x^{500}$. The term $x^{1001}$ is a single monomial. It only contributes to the coefficient of $x^{1001}$ in the expansion. Since $1001$ is much larger than $499$ or $500$, it has absolutely no impact on the coefficients we are interested in. It is essentially a 'distractor' term.

Q: How do I know when to use Pascal's Identity?

Pascal's Identity, $^{n}C_{r-1} + ^{n}C_r = ^{n+1}C_r$, is your best friend whenever you see the sum of two binomial coefficients that have the same upper index ($n$) and consecutive lower indices ($r-1$ and $r$). Whenever you see that pattern, immediately think of combining them into a single coefficient with an incremented upper index. It is a powerful shortcut that saves you from tedious calculations.

Question

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \dots + x^{1000}$ is :

Options:

(A) $^{1002}C_{500}$

(B) $^{1002}C_{501}$

(C) $^{1001}C_{501}$

(D) $^{1000}C_{501}$

Sigma Percentile · Accepted Answer

### The Beauty of Hidden Patterns

Imagine you are standing before a massive, intimidating wall of algebra. The expression $(1 + x)^{1000} + x(1 + x)^{999} + x^2(1 + x)^{998} + \dots + x^{1000}$ looks like a nightmare, doesn't it?

It is a long, sprawling series that seems to demand hours of tedious expansion. But here is the secret of JEE Advanced: math is rarely about brute force. It is about finding the hidden rhythm in the chaos.

If you look closely, you will see that each term is just the previous one multiplied by a specific factor. This is not just a random collection of terms; it is a **Geometric Progression (GP)**.

### Phase 1: The GP Realization

Let us define our parameters. The first term is $a = (1 + x)^{1000}$.

If you take the second term, $x(1 + x)^{999}$, and divide it by the first, you get the common ratio $r = \frac{x}{1+x}$. Because the powers of $x$ range from $0$ to $1000$, we have exactly $n = 1001$ terms.

Recognizing this structure is the 'Spark' that turns an impossible problem into a solvable one. We are no longer looking at a wall; we are looking at a staircase.

### Phase 2: The Algebraic Magic

Now, we apply the sum formula for a GP: $S = a \frac{1 - r^n}{1 - r}$. Substituting our values, we get:

$$S = (1 + x)^{1000} \cdot \frac{1 - (\frac{x}{1+x})^{1001}}{1 - \frac{x}{1+x}}$$

This looks messy, but watch what happens to the denominator. We have $1 - \frac{x}{1+x}$. If we find a common denominator, it becomes $\frac{(1+x) - x}{1+x}$, which simplifies beautifully to $\frac{1}{1+x}$.

Now, our expression is:

$$S = (1 + x)^{1000} \cdot \frac{1 - (\frac{x}{1+x})^{1001}}{\frac{1}{1+x}}$$

Dividing by a fraction is the same as multiplying by its reciprocal. So, the $\frac{1}{1+x}$ in the denominator flips up to become $(1+x)$, multiplying with our $(1+x)^{1000}$ to give us $(1+x)^{1001}$.

When we distribute this into the bracket, the $(1+x)^{1001}$ in the denominator of the second term cancels out perfectly. We are left with the elegant result:

$$S = (1 + x)^{1001} - x^{1001}$$

### Phase 3: The Binomial Extraction

All that complexity has vanished, leaving us with a simple binomial expression. We need the sum of the coefficients of $x^{499}$ and $x^{500}$.

The term $x^{1001}$ is irrelevant here, as it cannot produce these powers. We only need to look at $(1 + x)^{1001}$.

Using the Binomial Theorem, the general term is $^{n}C_{r}x^r$. For $x^{499}$, the coefficient is $^{1001}C_{499}$. For $x^{500}$, the coefficient is $^{1001}C_{500}$.

### Phase 4: The Elegant Finish

We are left with the sum $^{1001}C_{499} + ^{1001}C_{500}$. This is where we invoke **Pascal's Identity**: $^{n}C_{r-1} + ^{n}C_r = ^{n+1}C_r$.

With $n = 1001$ and $r = 500$, the sum becomes $^{1002}C_{500}$.

And just like that, the mountain is climbed. We didn't need to expand a single term. We used the structure of the series, the elegance of algebraic simplification, and the power of binomial identities to find the answer: **$^{1002}C_{500}$**.

Animated Solution for Mathematics - Binomial Theorem: The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is :

Select Answer:

Visualized Solution

Identifying the Series Structure

Defining GP Parameters

Applying GP Sum Formula

Simplifying the Denominator

Algebraic Reduction

Setting the Goal

Finding Coefficient of $x^{499}$

Finding Coefficient of $x^{500}$

Summing the Coefficients

Applying Pascal's Identity

Final Result

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Hidden Patterns

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; } The sum of the coefficients of x499 and x500 in (1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000 is :