If A denotes the sum of all the coefficients in the expansion of (1 - 3x + 10x²)ⁿ and B denotes the

Q: Wait, why do we set $x = 1$ to find the sum of coefficients? Shouldn't we be doing something with the binomial theorem?

That is a brilliant question! While you could expand the binomial using the general term formula, it would be an algebraic nightmare. By setting $x = 1$, we are essentially using the identity $P(1) = \sum a_i$. It is a powerful shortcut that bypasses the need for expansion entirely. Always look for the 'functional' approach before diving into the 'expansion' approach!

Q: What if the expression had a constant term without an $x$? Does the rule still hold?

Absolutely! If you have a polynomial like $P(x) = 5 + 2x + 3x^2$, the sum of coefficients is $5+2+3 = 10$. If you plug in $x=1$, you get $5 + 2(1) + 3(1)^2 = 10$. The constant term is just a coefficient of $x^0$, so it is naturally included in the sum.

Q: I got $A = 8^n$ and $B = 2^n$. How do I know which option is correct?

You are halfway there! You have $A = (2^3)^n$ and $B = 2^n$. Using the laws of exponents, $(2^n)^3$ is exactly $2^{3n}$. Since $B = 2^n$, replacing $2^n$ with $B$ gives you $A = B^3$. It is all about spotting the relationship between the bases.

Q: Does this work for negative coefficients too? Like the $-3x$ in the first expression?

Yes, it does! The rule $P(1)$ accounts for the sign of the coefficient automatically. When you substitute $x=1$, the term $-3x$ becomes $-3(1) = -3$. The sum of coefficients is the algebraic sum, meaning signs are preserved.

Q: Is there any scenario where this trick fails?

The trick is robust for any polynomial. However, be careful if the expression is not a polynomial (e.g., if it contains terms like $1/x$). As long as the expression is a polynomial in $x$, evaluating at $x=1$ will always yield the sum of its coefficients.

Question

If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3x + 10x^2)^n$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^2)^n$, then :

Options:

(A) $A = B^3$

(B) $3A = B$

(C) $B = A^3$

(D) $A = 3B$

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Polynomials

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to peel back the curtain on a problem that, at first glance, looks like a tedious exercise in binomial expansion.

You see the power $n$ and your mind might immediately jump to the Binomial Theorem, imagining pages of calculations. But stop. Take a breath.

In the world of competitive mathematics, the most elegant solution is often the one that requires the least amount of writing.

### The Golden Key: Evaluating at Unity

Consider any polynomial $P(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n$. If I asked you to find the sum of all its coefficients, $a_0 + a_1 + a_2 + \dots + a_n$, what would you do?

You could try to find each coefficient individually, but that is a trap. Instead, look at the structure. What happens if we set $x = 1$?

The expression becomes $P(1) = a_0 + a_1(1) + a_2(1)^2 + \dots + a_n(1)^n$. Suddenly, every power of $x$ vanishes, leaving behind exactly the sum we desire.

This is our **'Golden Rule'**: the sum of coefficients of any polynomial $P(x)$ is simply $P(1)$.

### Unmasking A and B

Let us apply this wisdom to our problem. We are given two expressions:

$$A = (1 - 3x + 10x^2)^n$$

$$B = (1 + x^2)^n$$

To find $A$, we evaluate the base at $x = 1$. It is almost too simple, isn't it? We substitute $x = 1$ into the base $(1 - 3x + 10x^2)$:

$$A = (1 - 3(1) + 10(1)^2)^n$$

Look at the arithmetic inside the bracket: $1 - 3 + 10$. That gives us $8$. So, $A = 8^n$.

Now, let us turn our attention to $B$. We evaluate $(1 + x^2)$ at $x = 1$:

$$B = (1 + (1)^2)^n = (1 + 1)^n = 2^n$$

### The Final Synthesis

Now we stand at the threshold of the solution. We have $A = 8^n$ and $B = 2^n$. We need to find the relationship between them.

This is where your intuition for numbers should kick in. We know that $8$ is not just a random number; it is a power of $2$. Specifically, $8 = 2^3$.

Let us rewrite $A$ using this knowledge:

$$A = (2^3)^n$$

By the laws of exponents, we know that $(a^m)^n = (a^n)^m$. Therefore, we can rearrange this as:

$$A = (2^n)^3$$

Since we already established that $B = 2^n$, we can perform a beautiful substitution. Replace $2^n$ with $B$, and what do we get?

**$$A = B^3$$**

### Reflection

Think about what we just achieved. We didn't need to expand the binomials. We didn't need to worry about the value of $n$.

We simply looked at the functional behavior of the polynomials at $x = 1$ and used the properties of exponents to bridge the gap. This is the essence of the JEE Advanced mindset: identifying the underlying structure, applying a powerful theorem, and letting the algebra resolve itself with grace.

Keep this **'Golden Rule'** in your toolkit—it will save you time and stress in many problems to come.

Animated Solution for Mathematics - Binomial Theorem: If $A$ denotes the sum of all the coefficients in the expansion of $(1 - 3 x + 10 x^{2})^{n}$ and $B$ denotes the sum of all the coefficients in the expansion of $(1 + x^{2})^{n}$ , then :

Select Answer:

Visualized Solution

The Problem Setup

The Golden Rule for Sum of Coefficients

Applying the Rule to $A$

Raw Substitution for $A$

Simplifying the Base of $A$

Converting $A$ to Base $2$

Applying the Rule to $B$

Calculating $B$

Finding the Relationship

The Final Result

The Sigma Insight: Properties of Binomial Coefficients

The Hidden Symmetry of Polynomials

The Golden Key

Unmasking A and B

The Final Synthesis

Reflection

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Select Answer:

Visualized Solution

The Problem Setup

The Golden Rule for Sum of Coefficients

Applying the Rule to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Raw Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Simplifying the Base of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Converting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A to Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Applying the Rule to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B