Finding Specific Coefficients Faster in Binomial Expansion

A Complete Guide for JEE Main with Shortcuts, Tricks & Previous Year Questions

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Introduction

In JEE Main, Binomial Theorem carries significant weightage with 1-2 questions appearing yearly. A recurring theme involves finding specific coefficients or terms independent of x. While the general term method works, time pressure demands faster alternatives.

This guide provides structured shortcuts to solve such problems efficiently.

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1. Core Concept: The General Term

For $(a + b)^n$, the general term is: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

This is fundamental, but deriving $r$ repeatedly is time-consuming. Let’s optimize.

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2. Shortcut for $(ax^p + bx^{-q})^n$

2.1 Formula for $r$

For the term containing $x^m$, use: $\boxed{r = \frac{np - m}{p + q}}$ Derivation: $T_{r+1} = \binom{n}{r} (ax^p)^{n-r} (bx^{-q})^r = \binom{n}{r} a^{n-r} b^r \, x^{p(n-r) - qr}$ Set the power of $x$ equal to $m$: $p(n-r) - qr = m \quad \Rightarrow \quad pn - pr - qr = m \quad \Rightarrow \quad r = \frac{np - m}{p + q}$

2.2 Term Independent of $x$ (Constant Term)

Set $m = 0$: $\boxed{r = \frac{np}{p + q}}$

2.3 Validity Check

• $r$ must be a non-negative integer.
• $0 \le r \le n$.
• If $r$ fails these conditions, the required term does not exist.

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3. Quick Reference Table

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4. Step-by-Step Application

Step 1: Identify parameters

Extract $n$, $p$, $q$, $a$, $b$ from $(ax^p + bx^{-q})^n$.

Step 2: Compute $r$

Use the appropriate formula.

Step 3: Validate $r$

Check if $r$ is a valid integer within range.

Step 4: Compute the coefficient

$\text{Coefficient} = \binom{n}{r} \cdot a^{n-r} \cdot b^r$ Don’t forget the signs if $b$ is negative.

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5. Solved Examples

Example 1: Independent term in $\left(x + \frac{1}{x}\right)^{10}$

• $n=10$, $p=1$, $q=1$
• $r = \frac{10 \times 1}{1+1} = 5$
• Term: $T_6 = \binom{10}{5} = 252$

Answer: 252

Example 2: Coefficient of $x^4$ in $\left(x^2 + \frac{1}{x}\right)^{10}$

• $n=10$, $p=2$, $q=1$, $m=4$
• $r = \frac{20 - 4}{3} = \frac{16}{3}$ (not integer)
• Term does not exist.

Example 3: Coefficient of $x^8$ in $\left(x^2 + \frac{1}{x}\right)^{10}$

• $r = \frac{20 - 8}{3} = 4$ ✓
• Coefficient = $\binom{10}{4} = 210$

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6. JEE Main Previous Year Questions (PYQs)

PYQ 1 (JEE Main 2013)

Problem: Independent term in $\left(\frac{x+1}{x^{2/3} - x^{1/3} + 1} - \frac{x-1}{x - x^{1/2}}\right)^{10}$.

Key Step: Simplify to $(x^{1/3} - x^{-1/2})^{10}$.

Solution:

• $n=10$, $p=\frac{1}{3}$, $q=\frac{1}{2}$
• $r = \frac{10 \times \frac{1}{3}}{\frac{1}{3} + \frac{1}{2}} = \frac{10/3}{5/6} = 4$
• Term = $\binom{10}{4} = 210$

Answer: 210

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PYQ 2 (JEE Main 2020)

Problem: Constant term in $\left(\sqrt{x} - \frac{k}{x^2}\right)^{10}$ is 405. Find $|k|$.

Solution:

• $n=10$, $p=\frac{1}{2}$, $q=2$
• $r = \frac{10 \times \frac{1}{2}}{\frac{1}{2} + 2} = \frac{5}{5/2} = 2$
• Constant term = $\binom{10}{2} k^2 = 45k^2 = 405$
• $k^2 = 9 \Rightarrow |k| = 3$

Answer: 3

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PYQ 3 (JEE Main 2021 September)

Problem: Independent term in $\left(\frac{x}{4} - \frac{12}{x^2}\right)^{12}$ is $\left(\frac{3^6}{4^4}\right)k$. Find $k$.

Solution:

• $n=12$, $p=1$, $q=2$
• $r = \frac{12}{3} = 4$
• Term = $\binom{12}{4} \cdot \left(\frac{1}{4}\right)^8 \cdot (-12)^4 = 495 \cdot \frac{12^4}{4^8}$
• Simplify to $\frac{729}{256} \cdot 55 = \frac{3^6}{4^4} \cdot 55$
• Thus $k = 55$

Answer: 55

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PYQ 4 (JEE Main 2022)

Problem: Find natural number $m$ if coefficient of $x$ in $\left(x^m + \frac{1}{x^2}\right)^{22}$ is 1540.

Solution:

• $n=22$, $q=2$, $m$ is unknown power in base.
• Coefficient corresponds to $\binom{22}{r} = 1540$.
• From values, $\binom{22}{3} = 1540$, so $r = 3$ or $19$.
• Using $r = \frac{22p - 1}{p + 2}$ (here $p = m$):
  • For $r=3$: $3 = \frac{22m-1}{m+2} \Rightarrow m = 7/19$ (invalid)
  • For $r=19$: $19 = \frac{22m-1}{m+2} \Rightarrow m = 13$

Answer: 13

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7. Extended Cases & Tricks

7.1 Expressions with More Than Two Terms

If the expression is like $(1 + 2x - 3x^3) \cdot (ax^p + bx^{-q})^n$:

• Find constant terms from the binomial expansion for powers that combine with the first factor to give $x^0$.
• For $(1)$: need $x^0$ from binomial.
• For $(2x)$: need $x^{-1}$ from binomial.
• For $(-3x^3)$: need $x^{-3}$ from binomial.
• Sum all valid contributions.

7.2 Simplification Before Expansion

Always check if the expression can be simplified algebraically before applying binomial theorem.

7.3 Maximum/Minimum of Independent Term

If the independent term contains a parameter $x$ (like in $\left(tx^{\alpha} + \frac{(1-x)^{\beta}}{t}\right)^n$), after finding the term as a function of $x$, use calculus (derivative = 0) to find extremum.

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8. Common Pitfalls & Tips

1. Sign Errors: Remember $b$ may be negative; include $(-1)^r$ when needed.
2. Fractional Powers: Handle $p, q$ carefully when they are fractions.
3. Invalid $r$: If $r$ is not an integer between 0 and $n$, the term does not exist.
4. Combining Factors: When a binomial is multiplied by another polynomial, consider all combinations that yield the target power.
5. Binomial Coefficient Calculation: Know common values: $\binom{n}{0}=1$, $\binom{n}{1}=n$, $\binom{n}{2}=\frac{n(n-1)}{2}$.

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9. Practice Problems

1. Independent term in $\left(x^2 - \frac{1}{x^3}\right)^{10}$

   • $n=10$, $p=2$, $q=3$
   • $r = \frac{20}{5} = 4$
   • Term = $\binom{10}{4} \cdot (1)^6 \cdot (-1)^4 = 210$

2. Coefficient of $x^5$ in $\left(x + \frac{2}{x^2}\right)^{11}$

   • $n=11$, $p=1$, $q=2$, $m=5$
   • $r = \frac{11 - 5}{3} = 2$
   • Coefficient = $\binom{11}{2} \cdot 2^2 = 55 \times 4 = 220$

3. Independent term in $\left(\sqrt{x} + \frac{1}{\sqrt[3]{x}}\right)^{10}$

   • $n=10$, $p=\frac{1}{2}$, $q=\frac{1}{3}$
   • $r = \frac{5}{1/2 + 1/3} = \frac{5}{5/6} = 6$
   • Term = $\binom{10}{6} = 210$

4. Relation between $a$ and $b$ if coefficients of $x^7$ in $\left(ax^2 + \frac{1}{bx}\right)^{11}$ and $x^{-7}$ in $\left(ax - \frac{1}{bx^2}\right)^{11}$ are equal.

   • For first: $p=2$, $q=1$, $m=7$ → $r_1 = \frac{22-7}{3} = 5$ Coefficient $C_1 = \binom{11}{5} a^6 b^{-5}$
   • For second: $p=1$, $q=2$, $m=-7$ → $r_2 = \frac{11 + 7}{3} = 6$ Coefficient $C_2 = \binom{11}{6} a^5 b^{-6} \cdot (-1)^6$
   • Equate $C_1 = C_2$: $\binom{11}{5} a^6 b^{-5} = \binom{11}{6} a^5 b^{-6} \quad \Rightarrow \quad a b = 1$

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10. Summary

• Shortcut: $r = \frac{np - m}{p+q}$ for $(ax^p + bx^{-q})^n$.
• Independent term: set $m=0$ → $r = \frac{np}{p+q}$.
• Always check validity of $r$.
• Simplify first, then apply the shortcut.
• For product expansions, consider each multiplicative term separately.

Mastering this approach will significantly reduce your problem-solving time in JEE Main.

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Best of luck for your JEE Main preparation! 🎯