Class 11 FSc Math Chapter 8 Binomial Theorem Solved Board Questions PDF | General Term, Middle Term & Term Independent of x | FBISE & Punjab Boards Study Guide

Theoretical Foundations & Core Concepts

The Binomial Theorem provides an algebraic expansion of powers of a binomial expression $(a+x)^n$ for any positive integer $n$. Prior to this algebraic tool, expanding higher-degree expressions like $(a+x)^7$ required tedious iterative multiplication. Formulated rigorously by Sir Isaac Newton, the theorem forms a foundational pillar for intermediate calculus, probability, and numerical analysis in engineering studies.

1. Statement of the Binomial Theorem

For any positive integer $n$ and real numbers $a$ and $x$:

$$(a+x)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}x + \binom{n}{2}a^{n-2}x^2 + \dots + \binom{n}{r}a^{n-r}x^r + \dots + \binom{n}{n}x^n$$

In summation notation, this is expressed compactly as:

$$(a+x)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} x^r$$

where the binomial coefficient is defined as $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

2. Key Algebraic Properties of the Expansion

• Total Number of Terms: The expansion of $(a+x)^n$ contains exactly $(n+1)$ terms.
• Sum of Indices: In every individual term, the sum of the exponents of $a$ and $x$ is equal to $n$, i.e., $(n-r) + r = n$.
• Symmetry of Coefficients: Binomial coefficients equidistant from the beginning and the end are equal because $\binom{n}{r} = \binom{n}{n-r}$.

3. General Term ($T_{r+1}$) Formula

The $(r+1)^{\text{th}}$ term in the expansion of $(a+x)^n$ is designated as the General Term. It allows direct evaluation of any specified term without expanding the entire polynomial:

$$T_{r+1} = \binom{n}{r} a^{n-r} x^r$$

Important Board Tip: Notice that for the $k^{\text{th}}$ term, we set $r = k - 1$. For instance, to find the $5^{\text{th}}$ term ($T_5$), we substitute $r = 4$.

4. Middle Term(s) Determination

The position of the middle term depends entirely on whether the exponent $n$ is even or odd:

• Case 1: When $n$ is EVEN
  The total number of terms $(n+1)$ is odd. Hence, there exists exactly one middle term situated at the position: $$\text{Position} = \left(\frac{n}{2} + 1\right)\text{th term}$$
• Case 2: When $n$ is ODD
  The total number of terms $(n+1)$ is even. Hence, there exist two middle terms situated at the positions: $$\text{Positions} = \left(\frac{n+1}{2}\right)\text{th term} \quad \text{and} \quad \left(\frac{n+3}{2}\right)\text{th term}$$

5. Term Independent of $x$ (Constant Term)

A term independent of $x$ is a term where the total exponent of $x$ is equal to $0$ (i.e., $x^0 = 1$). To calculate this term:

1. Write down the general term $T_{r+1}$ using the explicit values of $a$, $x$, and $n$.
2. Simplify and combine all powers of $x$ into a single exponent expression $x^{f(r)}$.
3. Set the exponent $f(r) = 0$ and solve for $r$.
4. Substitute $r$ back into $T_{r+1}$ to determine the value of the constant term.

Formula Summary & Quick Reference

Concept / Quantity	Mathematical Formula / Expression	Board Exam Application Tip
General Term ($T_{r+1}$)	$T_{r+1} = \binom{n}{r} a^{n-r} b^r$	Always set $r = \text{Term Position} - 1$. Keep track of negative signs in $b$.
Total Number of Terms	$N = n + 1$	If exponent is $12$, there are $13$ terms in the expansion.
Middle Term ($n$ is Even)	$T_{\frac{n}{2} + 1}$	Single middle term. Set $r = \frac{n}{2}$.
Middle Terms ($n$ is Odd)	$T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$	Two middle terms. Compute $r_1 = \frac{n-1}{2}$ and $r_2 = \frac{n+1}{2}$.
Term Independent of $x$	Exponent of $x = 0$	Solve $f(r) = 0$ for $r$. $r$ must be a non-negative integer ($r \in \mathbb{W}$).
Sum of Binomial Coefficients	$\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$	Frequently tested in FBISE & Punjab Board objective/MCQ sections.
Sum of Odd/Even Coefficients	$C_0 + C_2 + C_4 + \dots = C_1 + C_3 + C_5 + \dots = 2^{n-1}$	Useful for identity verification proofs in Board Long Questions.

Step-by-Step Solved Board Exam Questions

Question 1: Find the $5^{\text{th}}$ term in the expansion of $\left(2x - \frac{1}{x}\right)^{10}$. [BISE Lahore 2022, BISE Rawalpindi 2021]

Solution:

Step 1: Identify parameters from the given binomial expression.
Comparing $\left(2x - \frac{1}{x}\right)^{10}$ with $(a + b)^n$:

$$a = 2x, \quad b = -\frac{1}{x}, \quad n = 10$$

Step 2: State the general term formula.

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Step 3: Substitute $r = 4$ to find the $5^{\text{th}}$ term ($T_5$).

$$T_{4+1} = \binom{10}{4} (2x)^{10-4} \left(-\frac{1}{x}\right)^4$$

Step 4: Compute combinations and evaluate algebraic powers.

$$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$$ $$(2x)^6 = 64 x^6$$ $$\left(-\frac{1}{x}\right)^4 = \frac{1}{x^4}$$

Step 5: Multiply the simplified factors together.

$$T_5 = 210 \times (64 x^6) \times \frac{1}{x^4}$$ $$T_5 = 13440 x^{6-4}$$ $$T_5 = 13440 x^2$$

Final Answer: The $5^{\text{th}}$ term of the given expansion is $\mathbf{13440 x^2}$.

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Question 2: Find the term independent of $x$ in the expansion of $\left(x^2 + \frac{2}{x}\right)^9$. [FBISE 2019, BISE Gujranwala 2023]

Solution:

Step 1: Write down the General Term.
Here, $a = x^2$, $b = \frac{2}{x} = 2x^{-1}$, and $n = 9$.

$$T_{r+1} = \binom{9}{r} (x^2)^{9-r} (2x^{-1})^r$$

Step 2: Simplify exponents of $x$.

$$T_{r+1} = \binom{9}{r} x^{2(9-r)} \cdot 2^r \cdot x^{-r}$$ $$T_{r+1} = \binom{9}{r} 2^r \cdot x^{18 - 2r - r}$$ $$T_{r+1} = \binom{9}{r} 2^r \cdot x^{18 - 3r}$$

Step 3: Set exponent of $x$ equal to zero for term independent of $x$.

$$18 - 3r = 0 \implies 3r = 18 \implies r = 6$$

Step 4: Substitute $r = 6$ back into the simplified general term.

$$T_{6+1} = T_7 = \binom{9}{6} 2^6 x^0$$ $$\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$ $$2^6 = 64$$ $$T_7 = 84 \times 64 = 5376$$

Final Answer: The term independent of $x$ is the $7^{\text{th}}$ term, and its numerical value is $\mathbf{5376}$.

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Question 3: Find the middle term(s) in the expansion of $\left(\frac{3x}{2} - \frac{1}{3x}\right)^{11}$. [BISE Multan 2022, FBISE 2022]

Solution:

Step 1: Analyze the exponent $n$.
Here $n = 11$, which is odd. Therefore, the expansion contains $11 + 1 = 12$ terms, meaning there are two middle terms.

Step 2: Determine positions of the middle terms.

$$\text{First Middle Term Position} = \frac{n+1}{2} = \frac{11+1}{2} = 6^{\text{th}} \text{ term}$$ $$\text{Second Middle Term Position} = \frac{n+3}{2} = \frac{11+3}{2} = 7^{\text{th}} \text{ term}$$

Step 3: Calculate the $6^{\text{th}}$ term ($T_6$ where $r = 5$).

$$T_{5+1} = \binom{11}{5} \left(\frac{3x}{2}\right)^{11-5} \left(-\frac{1}{3x}\right)^5$$ $$\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$$ $$T_6 = 462 \cdot \left(\frac{3x}{2}\right)^6 \cdot \left(-\frac{1}{243 x^5}\right)$$ $$T_6 = 462 \cdot \frac{729 x^6}{64} \cdot \left(-\frac{1}{243 x^5}\right)$$ $$T_6 = 462 \cdot \frac{3 x^6}{64 x^5} \cdot (-1) = -\frac{1386}{64} x = -\frac{693}{32} x$$

Step 4: Calculate the $7^{\text{th}}$ term ($T_7$ where $r = 6$).

$$T_{6+1} = \binom{11}{6} \left(\frac{3x}{2}\right)^{11-6} \left(-\frac{1}{3x}\right)^6$$ $$\binom{11}{6} = \binom{11}{5} = 462$$ $$T_7 = 462 \cdot \left(\frac{3x}{2}\right)^5 \cdot \left(\frac{1}{729 x^6}\right)$$ $$T_7 = 462 \cdot \frac{243 x^5}{32} \cdot \frac{1}{729 x^6}$$ $$T_7 = 462 \cdot \frac{1}{32 \times 3 x} = \frac{462}{96 x} = \frac{77}{16x}$$

Final Answer: The two middle terms are $T_6 = \mathbf{-\frac{693}{32}x}$ and $T_7 = \mathbf{\frac{77}{16x}}$.

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Question 4: If $x^k$ occurs in the expansion of $\left(x + \frac{1}{x^2}\right)^n$, find the index formula for $r$ in terms of $n$ and $k$, and prove that $(n-k)$ must be a multiple of 3. [FBISE 2023, BISE Faisalabad 2020]

Solution:

Step 1: Formulate the General Term.
For $\left(x + x^{-2}\right)^n$, $a = x$, $b = x^{-2}$.

$$T_{r+1} = \binom{n}{r} (x)^{n-r} (x^{-2})^r$$ $$T_{r+1} = \binom{n}{r} x^{n-r-2r} = \binom{n}{r} x^{n-3r}$$

Step 2: Equate exponent of $x$ to $k$.
Since $x^k$ occurs in this term, we compare exponents of $x$:

$$n - 3r = k$$

Step 3: Solve for $r$ in terms of $n$ and $k$.

$$3r = n - k \implies r = \frac{n - k}{3}$$

Step 4: Deduce condition for existence.
In binomial expansions, $r$ represents a non-negative integer index ($r \in \{0, 1, 2, \dots, n\}$). For $r = \frac{n-k}{3}$ to be an integer, the numerator $(n-k)$ must be exactly divisible by $3$. Hence, $(n-k)$ must be a non-negative integer multiple of $3$. $\quad \blacksquare$

Interactive Practice Quiz (Clickable MCQs)

Test your conceptual understanding of Chapter 8 Binomial Theorem with board-pattern objective questions below:

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