If the constant term, in binomial expansion of (2x^r + 1/x²)¹⁰ is 180, then r is equal to ____.

Q: Why is the constant term the one where the power of $x$ is zero?

Think of $x^0$. Anything to the power of zero is $1$. So, if the variable part of your term is $x^0$, it becomes $1$, leaving only the coefficient. That's why we call it a constant term!

Q: What if I get a non-integer $k$?

In binomial expansion, $k$ represents the index of the term, so it must be a non-negative integer between $0$ and $n$. If you get a fraction, it's a sign that your algebraic setup for the exponent might be slightly off. Re-check your power laws!

Q: Why do we separate the coefficients from the variables?

It's a strategy of 'divide and conquer'. By separating them, you create two distinct equations: one for the variable (to find $k$) and one for the coefficient (to verify or solve for unknowns). It keeps your math clean and prevents errors.

Question

If the constant term, in binomial expansion of $(2x^r + \frac{1}{x^2})^{10}$ is 180, then $r$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex expression: $(2x^r + x^{-2})^{10}$. It looks like a chaotic mess of powers and variables, but in the world of JEE Advanced, we look for the underlying order.

The Binomial Theorem is our telescope, allowing us to zoom into any specific term without having to expand the entire expression. Our goal is to find the value of $r$ that makes the constant term equal to $180$.

### The Master Key: The General Term

To unlock this, we use the General Term formula: 
$$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$

This formula is the master key for all such problems. In our case, $n=10$, $a=2x^r$, and $b=x^{-2}$.

When we substitute these into the formula, we get:
$$T_{k+1} = \binom{10}{k} (2x^r)^{10-k} (x^{-2})^k$$

This is where the magic happens. We separate the expression into a numerical coefficient part, $\binom{10}{k} 2^{10-k}$, and a variable part, $x^{r(10-k) - 2k}$.

### The Power of Zero

Now, we address the core of the problem: the constant term. A constant term is, by definition, independent of $x$.

This means the exponent of $x$ must be zero. We set the exponent to zero:
$$r(10-k) - 2k = 0$$

This gives us a beautiful, simple relationship:
$$r = \frac{2k}{10-k}$$

This is our bridge. We know that if we find $k$, we find $r$.

### The Final Lock: Solving for $k$

We are told the constant term is $180$. We take our numerical coefficient part and set it equal to $180$:
$$\binom{10}{k} 2^{10-k} = 180$$

Since $k$ must be an integer between $0$ and $10$, we can test values. Let's try $k=8$:
$$\binom{10}{8} 2^{10-8} = \binom{10}{2} 2^2 = 45 	imes 4 = 180$$

It matches perfectly! With $k=8$ confirmed, we return to our bridge:
$$r = \frac{2(8)}{10-8} = \frac{16}{2} = 8$$

And there it is: **$r=8$**. The chaos has been tamed, and the order revealed.

Animated Solution for Mathematics - Binomial Theorem: If the constant term, in binomial expansion of $(2 x^{r} + \frac{1}{x ^{2}})^{10}$ is 180, then $r$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Identify the Binomial Expression

Recall the General Term Formula

Substitute Values into the Formula

Separate Constants and Variables

Apply the Constant Term Condition

Express r in terms of k

Set the Coefficient Equation

Solve for k by Testing Values

Calculate the Final Value of r

Conclusion and Key Takeaway

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Master Key

The Power of Zero

The Final Lock

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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; } If the constant term, in binomial expansion of (2xr+x21​)10 is 180, then r is equal to ____.