The coefficients of xᵖ and xᵗ in the expansion of (1 + x)ᵖ⁺ᵗ are (a) equal (b) equal with opposite s

Q: Wait, why is the general term $T_{r+1}$ and not $T_r$?

Great question! The index $r$ starts from 0. So the first term is $T_{0+1} = T_1$. It's just a convention to keep the index of the term consistent with the power of $x$.

Q: Does the order of $p$ and $q$ matter?

Not at all! The binomial coefficient ${}^{n}C_{r}$ is symmetric. Whether you choose $p$ or $q$, the underlying logic of combinations remains the same.

Q: Is this true for any binomial expansion?

Yes, the symmetry property ${}^{n}C_{r} = {}^{n}C_{n-r}$ is a fundamental identity for all binomial coefficients, regardless of the specific values of $n$, $p$, or $q$.

Question

The coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p+q}$ are

Options:

(A) equal

(B) equal with opposite signs

(C) reciprocals of each other

(D) none of these

Sigma Percentile · Accepted Answer

### The Beauty of Binomial Symmetry

Imagine you are standing before the expression $(1+x)^{p+q}$. At first glance, it looks like a simple algebraic construct, but beneath the surface lies a profound symmetry that governs the very nature of combinations.

Today, we are going to peel back the layers of this expression to understand why the coefficients of $x^p$ and $x^q$ are destined to be identical.

### The Microscope: The General Term

To analyze any binomial expansion, we need a tool that allows us to zoom in on any specific term. That tool is the general term formula.

For an expansion of $(1+x)^n$, the general term, $T_{r+1}$, is given by:

$$T_{r+1} = {}^{n}C_{r} x^{r}$$

Think of this as our microscope. By choosing the right value for $r$, we can isolate any coefficient we desire. In our specific case, the total power is $n = p+q$.

So, our microscope becomes:

$$T_{r+1} = {}^{p+q}C_{r} x^{r}$$

This is our master key.

### The Hunt for Coefficients

Now, let's hunt for our targets. We want the coefficient of $x^p$. Looking at our formula, we need the power of $x$ to be $p$.

So, we simply set $r = p$. When we do that, the term becomes:

$$T_{p+1} = {}^{p+q}C_{p} x^{p}$$

The coefficient is clearly ${}^{p+q}C_{p}$.

Similarly, to find the coefficient of $x^q$, we set $r = q$. The term becomes:

$$T_{q+1} = {}^{p+q}C_{q} x^{q}$$

The coefficient is ${}^{p+q}C_{q}$. Now, we have our two coefficients: ${}^{p+q}C_{p}$ and ${}^{p+q}C_{q}$.

### The Mirror: The Symmetry Property

This is where the magic happens. We invoke the fundamental symmetry property of binomial coefficients:

$${}^{n}C_{r} = {}^{n}C_{n-r}$$

Why is this true? Think about it logically: choosing $r$ items from a group of $n$ is exactly the same as deciding which $n-r$ items to leave behind. The number of ways to do both is identical.

Let's apply this to our first coefficient, ${}^{p+q}C_{p}$. Here, $n = p+q$ and $r = p$. According to the property, ${}^{p+q}C_{p}$ must be equal to ${}^{p+q}C_{(p+q)-p}$.

### The Final Revelation

Now, let's simplify that lower index. We have $(p+q) - p$. The $p$ and $-p$ cancel each other out perfectly, leaving us with just $q$.

So, ${}^{p+q}C_{p}$ simplifies exactly to ${}^{p+q}C_{q}$. We have arrived at our destination!

The coefficient of $x^p$ is exactly the same as the coefficient of $x^q$. They are **perfectly equal**. This is the elegance of mathematics—a seemingly complex comparison resolved by a simple, beautiful symmetry.

Animated Solution for Mathematics - Binomial Theorem: The coefficients of $x^{p}$ and $x^{q}$ in the expansion of $(1 + x)^{p + q}$ are

Select Answer:

Visualized Solution

The Binomial Expression

General Term Formula

General Term for Our Expression

Coefficient of $x^{p}$

Coefficient of $x^{q}$

The Symmetry Property

Applying Symmetry

Simplifying the Expression

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

The Beauty of Binomial Symmetry

The Microscope

The Hunt for Coefficients

The Mirror

The Final Revelation

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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 coefficients of xp and xq in the expansion of (1+x)p+q are