The sum, of the coefficients of the first 50 terms in the binomial expansion of (1 - x)¹⁰⁰, is equal

Q: Wait, why is the 50th term corresponding to $r=49$ and not $r=50$?

This is a classic trap! In the binomial expansion of $(1-x)^{100}$, the terms are indexed starting from $r=0$. The first term is for $r=0$, the second for $r=1$, and so on. Following this pattern, the 50th term corresponds to $r=49$. Always count your indices carefully!

Q: Why does the total sum of coefficients equal $0$ when we set $x=1$?

Think about the identity $(1-x)^{100} = \sum_{r=0}^{100} (-1)^r \binom{100}{r} x^r$. If you set $x=1$, the left side becomes $(1-1)^{100} = 0^{100} = 0$. The right side becomes the sum of all coefficients $\binom{100}{0} - \binom{100}{1} + \binom{100}{2} - \dots + \binom{100}{100}$. Thus, the sum must be zero.

Q: How can we be sure that the sum of the last 50 terms is equal to $S$?

By using the symmetry property $\binom{n}{r} = \binom{n}{n-r}$, we can map each term in the second half of the expansion to a corresponding term in the first half. Because the signs alternate, the symmetry ensures that the second half mirrors the first half, allowing us to group them together.

Question

The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$, is equal to

Options:

(A) $^{101}C_{50}$

(B) $^{99}C_{49}$

(C) $-^{101}C_{50}$

(D) $-^{99}C_{49}$

Sigma Percentile · Accepted Answer

### The Binomial Landscape: A Journey into Symmetry

Welcome, fellow traveler of the mathematical realm. Today, we stand before a problem that might look like a daunting mountain of coefficients, but it is a landscape of profound elegance.

We are looking at the binomial expansion of $(1-x)^{100}$ and seeking the sum of its first 50 terms. Let us embark on this journey together.

### Phase 1: The Alternating Dance

First, let us visualize the expansion. When we expand $(1-x)^{100}$, we are dealing with a rhythmic, alternating dance of coefficients.

The general form is:

$$ \binom{100}{0} - \binom{100}{1}x + \binom{100}{2}x^2 - \dots + \binom{100}{100}x^{100} $$

Notice the signs? They alternate: positive, negative, positive, negative. This is the heartbeat of our problem.

We are interested in the sum of the first 50 terms. Since we start at $r=0$, the 50th term is at $r=49$. Let us call this sum $S$:

$$ S = \binom{100}{0} - \binom{100}{1} + \binom{100}{2} - \dots - \binom{100}{49} $$

### Phase 2: The Zero Sum Revelation

Now, here is the masterstroke. What happens if we look at the entire expansion?

If we set $x=1$, the expression $(1-x)^{100}$ becomes $(1-1)^{100}$, which is simply $0$. This means the sum of all coefficients, from $r=0$ to $r=100$, must be exactly zero.

This is our foundation:

$$ \binom{100}{0} - \binom{100}{1} + \dots + \binom{100}{100} = 0 $$

### Phase 3: The Symmetry of the Mirror

We have 101 terms in total. We have our first 50 terms (the sum $S$), the middle term $\binom{100}{50}$, and the last 50 terms.

By the beautiful symmetry property of binomial coefficients, $\binom{n}{r} = \binom{n}{n-r}$, we know that $\binom{100}{100} = \binom{100}{0}$, $\binom{100}{99} = \binom{100}{1}$, and so on. When we apply this to the last 50 terms, we find that they mirror the first 50 terms perfectly.

The entire sum becomes:

$$ S + \binom{100}{50} + S = 0 $$

This simplifies to $2S + \binom{100}{50} = 0$, or:

$$ S = -\frac{1}{2} \binom{100}{50} $$

### Phase 4: The Final Polish

We are almost there! We need to express the result in terms of $\binom{99}{49}$.

Using the index reduction formula, $\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$, we can rewrite $\binom{100}{50}$ as:

$$ \binom{100}{50} = \frac{100}{50} \binom{99}{49} = 2 \binom{99}{49} $$

Substituting this back into our equation for $S$, we get:

$$ S = -\frac{1}{2} 	imes 2 \binom{99}{49} $$

This simplifies beautifully to our final result:

$$ \mathbf{S = -\binom{99}{49}} $$

The complexity melts away, leaving behind a simple, elegant result. You have navigated the symmetry, mastered the index reduction, and conquered the binomial expansion.

Animated Solution for Mathematics - Binomial Theorem: The sum, of the coefficients of the first 50 terms in the binomial expansion of $(1 - x)^{100}$ , is equal to

Select Answer:

Visualized Solution

Expansion of $(1 - x)^{100}$

Identifying the Coefficients

Defining the Target Sum $S$

The Total Alternating Sum Property

Symmetry of Binomial Coefficients

Splitting the Total Sum

Solving for $S$

Applying Index Reduction

Final Calculation and Result

Key Takeaways

The Sigma Insight: Properties of Binomial Coefficients

The Binomial Landscape

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 the first 50 terms in the binomial expansion of (1−x)100, is equal to