How do I decide whether to use differentiation or integration for Binomial Series problems?

Examine the variable coefficient multiplying $\binom{n}{r}$. If the coefficient $r$ is in the numerator (e.g., $\sum r \binom{n}{r}$), differentiate $(1+x)^n$ with respect to $x$. If $r+1$ is in the denominator (e.g., $\sum \frac{1}{r+1} \binom{n}{r}$), integrate $(1+x)^n$ with respect to $x$ over the interval $[0, 1]$.

Why do remainder problems using Binomial Expansion sometimes yield negative remainders, and how do I fix them?

When expanding expressions like $(M - 1)^n$, the last term can evaluate to $-1$, producing a result like $k m - 1$. Because remainders must strictly satisfy $0 \le R < m$, add and subtract $m$: $k m - 1 = (k-1)m + (m - 1)$, giving the correct positive remainder $m - 1$.

What is the fastest strategy to solve term independent of $x$ in complex algebraic expressions?

Express the general term $T_{r+1}$ in the simplified form $C(r) \cdot x^{g(r)}$. Set $g(r) = 0$ to solve for $r$. Verify that $r$ is a non-negative integer within $0 \le r \le n$; if $r$ is fractional or negative, no term independent of $x$ exists.

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Q: What is the fastest strategy to solve term independent of $x$ in complex algebraic expressions?

Express the general term $T_{r+1}$ in the simplified form $C(r) \cdot x^{g(r)}$. Set $g(r) = 0$ to solve for $r$. Verify that $r$ is a non-negative integer within $0 \le r \le n$; if $r$ is fractional or negative, no term independent of $x$ exists.

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Solved High-Yield Sample Questions

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Question #1JEE Main 2024

Find the remainder when $7^{103}$ is divided by $25$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Rewrite $7^{103} = 7 \cdot 7^{102} = 7 \cdot (49)^{51} = 7 \cdot (50 - 1)^{51}$ .

Step 2

Expand $(50 - 1)^{51}$ using Binomial Theorem: $(50 - 1)^{51} = \sum_{r=0}^{51} \binom{51}{r} 50^{51-r} (-1)^r = 50 k - 1$ , where $k$ is an integer.

Step 3

Multiply by $7$ : $7^{103} = 7(50k - 1) = 350k - 7 = 25(14k) - 7 = 25(14k - 1) + (25 - 7) = 25 m + 18$ . Thus, the remainder is $18$ .

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Question #2JEE Main 2023

If the coefficient of $x^7$ in $\left(a x^2 + \frac{1}{b x}\right)^{11}$ equals the coefficient of $x^{-7}$ in $\left(a x - \frac{1}{b x^2}\right)^{11}$ , then the relation between $a$ and $b$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

For $\left(a x^2 + \frac{1}{b x}\right)^{11}$ , general term $T_{r+1} = \binom{11}{r} (a x^2)^{11-r} \left(\frac{1}{b x}\right)^r = \binom{11}{r} a^{11-r} b^{-r} x^{22 - 3r}$ . For $x^7$ , set $22 - 3r = 7 \implies r = 5$ . Coefficient $= \binom{11}{5} a^6 b^{-5}$ .

Step 2

For $\left(a x - \frac{1}{b x^2}\right)^{11}$ , general term $T_{k+1} = \binom{11}{k} (a x)^{11-k} \left(-\frac{1}{b x^2}\right)^k = \binom{11}{k} (-1)^k a^{11-k} b^{-k} x^{11 - 3k}$ . For $x^{-7}$ , set $11 - 3k = -7 \implies k = 6$ . Coefficient $= \binom{11}{6} (-1)^6 a^5 b^{-6} = \binom{11}{6} a^5 b^{-6}$ .

Step 3

Equating both coefficients: $\binom{11}{5} a^6 b^{-5} = \binom{11}{6} a^5 b^{-6}$ . Since $\binom{11}{5} = \binom{11}{6}$ , we obtain $a^6 b^{-5} = a^5 b^{-6} \implies a b = 1$ .

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Mastering Binomial Theorem: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

General Term in Positive Integral Index

Middle Term Determination

Binomial Expansion for Any Index

Sum of Binomial Coefficients

Calculus-Based Coefficient Summation

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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