The coefficient of xⁿ in expansion of (1 + x)(1 - x)ⁿ is (a) (-1)ⁿ⁻¹n (b) (-1)ⁿ(1 - n) (c) (-1)ⁿ⁻¹(n

Q: Why can't we just expand the whole thing $(1 + x)(1 - x)^n$?

Efficiency is the soul of JEE preparation. While you *could* expand the entire expression, it is a massive trap that wastes precious time. By using the distributive property, we isolate only the terms that contribute to $x^n$, saving you from unnecessary algebraic clutter.

Q: Why do we look for $x^{n-1}$ in the second part?

This is a classic point of confusion! In the second part, we have $x \cdot (1 - x)^n$. Since there is already an $x$ outside the bracket, it acts as a multiplier. If we need an overall power of $x^n$, we only need the expansion of $(1 - x)^n$ to provide $x^{n-1}$, because $x \cdot x^{n-1} = x^n$.

Q: How do we handle the negative sign in $(-1)^{n-1}$?

Don't let the negative signs intimidate you! Use the laws of exponents: $(-1)^{n-1} = \frac{(-1)^n}{-1} = -(-1)^n$. By converting both terms to have the same $(-1)^n$ factor, you can easily combine them and factor out the common term.

Question

The coefficient of $x^n$ in expansion of $(1 + x)(1 - x)^n$ is

Options:

(A) $(-1)^{n-1}n$

(B) $(-1)^n(1 - n)$

(C) $(-1)^{n-1}(n - 1)^2$

(D) $(n - 1)$

Sigma Percentile · Accepted Answer

### The Art of Decomposition: Mastering the Binomial

Welcome, future engineers! Today, we are going to dissect a problem that, at first glance, might look like a tedious exercise in expansion. We are tasked with finding the coefficient of $x^n$ in the expansion of $(1 + x)(1 - x)^n$.

Many students fall into the trap of trying to expand the entire expression, but I want you to pause. In the world of JEE, the most elegant path is rarely the longest one. Let us embark on a journey to solve this with precision and grace.

### Phase 1: The Distributive Strategy

Imagine you are standing before a locked door. You don't need to break the whole wall; you just need the right key. Our expression is $(1 + x)(1 - x)^n$.

If we try to expand $(1 - x)^n$ fully, we are looking at a long string of terms. Instead, let us use the distributive property. We can split our expression into two manageable pieces:

$$1 \cdot (1 - x)^n + x \cdot (1 - x)^n$$

By doing this, we have transformed one complex problem into two smaller, simpler ones. We just need to find the coefficient of $x^n$ in each part and sum them up. It is a classic 'divide and conquer' strategy.

### Phase 2: The General Term - Your Mathematical Compass

To navigate the expansion of $(1 - x)^n$, we need our trusty compass: the general term formula. Recall that for any binomial $(a + b)^n$, the general term is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

In our case, $a = 1$ and $b = -x$. Thus, the general term becomes:

$$T_{r+1} = \binom{n}{r} (1)^{n-r} (-x)^r = \binom{n}{r} (-1)^r x^r$$

This formula is your master key. It tells us exactly what the coefficient of $x^r$ is for any value of $r$. Now, we are ready to tackle the two-front war.

### Phase 3: The Two-Front War

Let us look at the first part: $1 \cdot (1 - x)^n$. We need the coefficient of $x^n$. Looking at our general term $\binom{n}{r} (-1)^r x^r$, we clearly need $r = n$.

Substituting this, we get:

$$	ext{Coefficient}_1 = \binom{n}{n} (-1)^n$$

Now, consider the second part: $x \cdot (1 - x)^n$. This is where many students stumble. We have an $x$ outside the bracket, which is already contributing one power of $x$.

Therefore, to get an overall $x^n$, we only need the expansion of $(1 - x)^n$ to provide $x^{n-1}$. So, we set $r = n - 1$. Substituting this into our general term, we get:

$$	ext{Coefficient}_2 = \binom{n}{n-1} (-1)^{n-1}$$

### Phase 4: The Final Synthesis

We are almost there! The total coefficient is the sum of these two parts:

$$	ext{Total} = \binom{n}{n} (-1)^n + \binom{n}{n-1} (-1)^{n-1}$$

Now, let us use our knowledge of combinations. We know that $\binom{n}{n} = 1$ and $\binom{n}{n-1} = n$. Substituting these values, the expression simplifies to:

$$1 \cdot (-1)^n + n \cdot (-1)^{n-1}$$

To combine these, we need the powers of $-1$ to match. We know that $(-1)^{n-1} = -(-1)^n$. Let us substitute this back into our equation:

$$(-1)^n + n \cdot (-(-1)^n) = (-1)^n - n(-1)^n$$

Finally, we factor out the common term $(-1)^n$:

$$(-1)^n (1 - n)$$

And there it is! The elegance of mathematics reveals the answer. We didn't need to expand everything; we just needed to understand the structure. The final result is **$(-1)^n (1 - n)$**.

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is

Select Answer:

Visualized Solution

Identify the Expression

Expand using Distributive Property

Define the General Term

Coefficient from the First Part

Coefficient from the Second Part

Combine the Coefficients

Evaluate the Combinations

Align the Powers of Negative One

Factorize and Finalize

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Art of Decomposition

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 coefficient of xn in expansion of (1+x)(1−x)n is