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
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.
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 $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
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 $r = 4$.
The position of the middle term depends entirely on whether the exponent $n$ is even or odd:
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:
| 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. |
Solution:
Step 1: Identify parameters from the given binomial expression.
Comparing $\left(2x - \frac{1}{x}\right)^{10}$ with $(a + b)^n$:
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 $\mathbf{13440 x^2}$.
Solution:
Step 1: Write down the General Term.
Here, $a = x^2$, $b = \frac{2}{x} = 2x^{-1}$, and $n = 9$.
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 $\mathbf{5376}$.
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 $T_7 = \mathbf{\frac{77}{16x}}$.
Solution:
Step 1: Formulate the General Term.
For $\left(x + x^{-2}\right)^n$, $a = x$, $b = x^{-2}$.
Step 2: Equate exponent of $x$ to $k$.
Since $x^k$ occurs in this term, we compare exponents of $x$:
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 $3$. $\quad \blacksquare$
Test your conceptual understanding of Chapter 8 Binomial Theorem with board-pattern objective questions below:
Q1: What is the total number of terms in the expansion of $(2x - 3y)^{15}$?
Q2: What is the middle term position in the expansion of $(a + b)^{14}$?
Q3: The sum of all binomial coefficients $\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}$ is equal to:
Q4: In the general term formula $T_{r+1} = \binom{n}{r} a^{n-r} b^r$, what value of $r$ corresponds to finding the $7^{\text{th}}$ term?
Q5: What is the term independent of $x$ in the expansion of $\left(x + \frac{1}{x}\right)^4$?
Q6: If $n$ is an odd positive integer, how many middle terms exist in the expansion of $(a+x)^n$?
Q7: The coefficient of the $x^n$ term in the expansion of $(1+x)^{2n}$ is:
Q8: The sum of coefficients of odd terms $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots$ equals:
Answer: Express the general term $T_{r+1}$ and collect all exponents of $x$ into a single power equation $x^{f(r)}$. Set $f(r) = 0$ and solve for $r$. If $r$ turns out to be a non-negative integer ($r \in \{0, 1, 2, \dots, n\}$), then a constant term exists in the expansion. If $r$ is fractional or negative, no term independent of $x$ exists in that particular expansion.
Answer: The three most frequent errors reported by board examiners are:
1) Sign Errors: Forgetting to include the negative sign when $b$ is negative, such as taking $b = \frac{1}{x}$ instead of $b = -\frac{1}{x}$ in $\left(2x - \frac{1}{x}\right)^n$.
2) Incorrect Index $r$: Writing $r = 5$ when calculating the $5^{\text{th}}$ term instead of using $r = 4$.
3) Middle Term Confusion: Calculating only one middle term when $n$ is odd instead of evaluating both middle terms.
Answer:
• For positive integers $n$ (Section 8.2), the expansion contains finite $(n+1)$ terms, requires no restriction on $x$, and uses factorial binomial coefficients $\binom{n}{r}$.
• For fractional or negative $n$ (Section 8.3 / Binomial Series), the expansion contains infinite terms, is valid only if $|x| < 1$, and uses open product formulas for coefficients.