The sum of the co-efficients of all odd degree terms in the expansion of (x + √(x³ - 1))⁵ + (x - √(x

Q: Wait, why do we only keep the even terms in the binomial expansion?

Great observation! When we expand $(x + a)^n$ and $(x - a)^n$, the terms with odd powers of $a$ have opposite signs. For example, in $(x+a)^n$, the term $\binom{n}{1} x^{n-1} a^1$ is positive, but in $(x-a)^n$, it becomes $-\binom{n}{1} x^{n-1} a^1$. When you add the two expressions, these terms cancel out perfectly, leaving only the even powers of $a$ (like $a^0, a^2, a^4$), which are doubled.

Q: The question asks for the sum of coefficients of *odd degree terms*. Does that mean I should look at the powers of $x$ or the powers of the whole term?

The question refers to the degree of the polynomial in terms of $x$. After you expand and simplify the expression into a polynomial in $x$, you look at the exponent of each term. If the exponent is odd (like $x^1, x^3, x^5, x^7$), that term is an 'odd degree term'. You then sum the coefficients of these specific terms.

Q: I got 1, but the answer is 2. Did I miss something?

You are very close! Remember the identity $(x + a)^n + (x - a)^n = 2 \times [\dots]$. That factor of $2$ outside the bracket is crucial. You likely calculated the sum of the coefficients *inside* the bracket correctly as $1$, but forgot to multiply the entire result by $2$ at the very end. Always keep track of those constants!

Question

The sum of the co-efficients of all odd degree terms in the expansion of $(x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5, (x > 1)$ is :

Options:

(A) 2

(B) -1

(C) 0

(D) 1

Sigma Percentile · Accepted Answer

Welcome, future engineer. Today, we stand before a problem that, at first glance, seems designed to test your patience rather than your intellect. You see an expression like $(x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$ and your mind immediately jumps to the worst-case scenario: expanding a fifth-degree binomial with a square root term.

But stop. Breathe. In the world of JEE Advanced, the most intimidating problems are often the ones that hide the most elegant shortcuts. This is not a test of your ability to perform tedious algebra; it is a test of your ability to recognize symmetry.

### The Symmetry of Conjugates

When you see an expression of the form $(x + a)^n + (x - a)^n$, you are not looking at a random collection of terms. You are looking at a beautiful, balanced structure.

When we add $(x + a)^n$ and $(x - a)^n$, the terms with odd powers of $a$ involve a subtraction that results in cancellation. Specifically, the terms with $a^1, a^3, a^5$ will vanish. We are left only with the even powers of $a$: $a^0, a^2, a^4$.

This is our first breakthrough. We have reduced a potentially massive expansion into a manageable sum of three terms.

### The Expansion

Now, let us apply this to our specific case where $n=5$ and $a = \sqrt{x^3 - 1}$. The identity tells us that:

$$ (x + a)^5 + (x - a)^5 = 2 \left[ \binom{5}{0} x^5 + \binom{5}{2} x^3 a^2 + \binom{5}{4} x a^4 ight] $$

Notice how the powers of $x$ decrease by two while the powers of $a$ increase by two. This is the heartbeat of the binomial theorem.

Calculating the coefficients is straightforward: $\binom{5}{0} = 1$, $\binom{5}{2} = 10$, and $\binom{5}{4} = 5$. Our expression becomes:

$$ 2 \left[ x^5 + 10x^3 a^2 + 5x a^4 ight] $$

This is where the magic happens. We substitute $a^2 = x^3 - 1$ and $a^4 = (x^3 - 1)^2$. The square roots are gone, and we are now dealing with a standard polynomial.

### The Final Hunt

We expand $(x^3 - 1)^2$ to get $x^6 - 2x^3 + 1$. Substituting this back, we get:

$$ 2 \left[ x^5 + 10x^3(x^3 - 1) + 5x(x^6 - 2x^3 + 1) ight] $$

Distributing the terms, we get:

$$ 2 \left[ x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x ight] $$

Now, we identify the terms: $5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x$.

The problem asks for the sum of the coefficients of the resulting polynomial. Summing the coefficients $5 + 10 + 1 - 10 - 10 + 5$ gives $1$.

Finally, don't forget the factor of $2$ outside! The total sum is **$2 	imes 1 = 2$**.

You see? The monster was just a paper tiger. Keep this perspective, and you will conquer any problem.

Animated Solution for Mathematics - Binomial Theorem: The sum of the co-efficients of all odd degree terms in the expansion of $(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}, (x > 1)$ is :

Select Answer:

Visualized Solution

Identify the Binomial Pattern

Apply the Expansion Identity

Substitute $n = 5$ into the Formula

Calculate Binomial Coefficients

Substitute the Value of $a$

Expand the Squared Term

Distribute and Expand All Terms

Identify Odd Degree Terms

Extract Coefficients of Odd Degree Terms

Calculate the Final Sum

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Symmetry of Conjugates

The Expansion

The Final Hunt

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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 co-efficients of all odd degree terms in the expansion of (x+x3−1​)5+(x−x3−1​)5,(x>1) is :