This MCQ module is based on: 7.1 Introduction
TOPIC 21 OF 45
7.1 Introduction
🎓 Class 11
Mathematics
CBSE
Theory
Ch 7 — Binomial Theorem
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: 7.1 Introduction
Targeting Class 11 level in Algebra, with Advanced difficulty.
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7.1 Introduction
You already know the identities \((a+b)^2=a^2+2ab+b^2\) and \((a+b)^3=a^3+3a^2b+3ab^2+b^3\) from earlier classes. Expanding \((a+b)^4\) directly takes patience; \((a+b)^{20}\) by hand is hopeless. The Binomial Theorem? gives a clean closed-form expansion of \((a+b)^n\) for any positive integer \(n\), expressing it as a sum of \(n+1\) terms with elegantly patterned coefficients — the binomial coefficients? arranged in Pascal's Triangle?.
Blaise Pascal
1623 – 1662
French mathematician, physicist and theologian. His Traité du Triangle Arithmétique (1665) gave the most systematic treatment of the binomial-coefficient triangle then known in Europe, and used it to solve probability problems posed by gambler Chevalier de Méré — birthing the modern theory of probability. He also invented one of the first mechanical calculators.
7.2 Binomial Theorem for Positive Integral Indices
Let us look at the patterns when we expand \((a+b)^n\) for small \(n\):
\[\begin{aligned} (a+b)^0 &= 1\\ (a+b)^1 &= a+b\\ (a+b)^2 &= a^2+2ab+b^2\\ (a+b)^3 &= a^3+3a^2b+3ab^2+b^3\\ (a+b)^4 &= a^4+4a^3b+6a^2b^2+4ab^3+b^4 \end{aligned}\]Three patterns leap out:
- The total number of terms in \((a+b)^n\) is \(n+1\).
- Powers of \(a\) decrease from \(n\) to \(0\); powers of \(b\) increase from \(0\) to \(n\). Each term has total degree \(n\).
- The coefficients form a beautiful triangle.
Pascal's Triangle
Arrange the coefficients in a triangle, one row per power. Each entry equals the sum of the two entries directly above it (the boundary 1's are stipulated):
Row 0: 1
Row 1: 11
Row 2: 121
Row 3: 1331
Row 4: 14641
Row 5: 15101051
Row 6: 1615201561
Fig 7.1 / 7.2: Pascal's Triangle. Highlighted: \(\binom{4}{2}=6\), the third entry of row 4 — equals \(\binom{3}{1}+\binom{3}{2}=3+3\) (the two entries above it).
Each entry is exactly the binomial coefficient \(\binom{n}{k}=\,^nC_k=\dfrac{n!}{k!(n-k)!}\). The recurrence "two entries above sum to the entry below" is the famous Pascal identity:
Pascal's identity
\[\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}\quad\text{for } 1\le k\le n.\]
Statement of the Binomial Theorem
Binomial Theorem (positive integral index)
For any positive integer \(n\) and any real (or complex) numbers \(a,b\),
\[\boxed{\;(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{\,n-k}b^{k}\;}\]
\[=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n-1}ab^{n-1}+\binom{n}{n}b^{n}.\]
Proof by mathematical induction
Proof
Base \(n=1\): \((a+b)^1=a+b=\binom{1}{0}a+\binom{1}{1}b\). ✓Induction step: Assume the theorem for some \(n=m\): \[(a+b)^m=\sum_{k=0}^{m}\binom{m}{k}a^{m-k}b^{k}.\] Multiply both sides by \(a+b\): \[(a+b)^{m+1}=(a+b)\sum_{k=0}^{m}\binom{m}{k}a^{m-k}b^{k}=\sum_{k=0}^{m}\binom{m}{k}a^{m+1-k}b^{k}+\sum_{k=0}^{m}\binom{m}{k}a^{m-k}b^{k+1}.\] Re-index the second sum (\(k\to k-1\)) and combine like terms (the term with \(a^{m+1-k}b^{k}\)): \[=\binom{m}{0}a^{m+1}+\sum_{k=1}^{m}\!\!\left[\binom{m}{k}+\binom{m}{k-1}\right]\!a^{m+1-k}b^{k}+\binom{m}{m}b^{m+1}.\] By Pascal's identity, the bracket equals \(\binom{m+1}{k}\), and \(\binom{m}{0}=\binom{m+1}{0}=1\), \(\binom{m}{m}=\binom{m+1}{m+1}=1\). Hence \[(a+b)^{m+1}=\sum_{k=0}^{m+1}\binom{m+1}{k}a^{m+1-k}b^{k}.\] This is the theorem for \(n=m+1\). By induction the theorem holds for every positive integer \(n\). \(\square\)
Interactive: Expand \((a+b)^n\) for any n ≤ 12
Move the slider to choose \(n\). The simulation prints out the full binomial expansion and the corresponding row of Pascal's Triangle.
5
Row 5: 1, 5, 10, 10, 5, 1
(a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵
Worked Examples
Example 1. Expand \(\left(x^2+\dfrac{3}{x}\right)^4\) using the binomial theorem.
With \(a=x^2,\ b=3/x,\ n=4\):
\[\sum_{k=0}^{4}\binom{4}{k}(x^2)^{4-k}\!\left(\dfrac{3}{x}\right)^{\!k}=\sum_{k=0}^{4}\binom{4}{k}3^k x^{8-3k}.\]
Term-by-term: \(\binom{4}{0}x^8+\binom{4}{1}\cdot 3x^5+\binom{4}{2}\cdot 9x^2+\binom{4}{3}\cdot 27\cdot x^{-1}+\binom{4}{4}\cdot 81 x^{-4}\)
\(=x^8+12x^5+54x^2+\dfrac{108}{x}+\dfrac{81}{x^4}\).
Example 2. Compute \((98)^5\) using the binomial theorem.
\(98=100-2\). Apply theorem with \(a=100,\ b=-2,\ n=5\):
\[(100-2)^5=\binom{5}{0}100^5-\binom{5}{1}100^4\cdot 2+\binom{5}{2}100^3\cdot 4-\binom{5}{3}100^2\cdot 8+\binom{5}{4}100\cdot 16-\binom{5}{5}32.\]
\(=10\,000\,000\,000-1\,000\,000\,000+40\,000\,000-800\,000+8\,000-32=9\,039\,207\,968\).
Activity: Build Pascal's Triangle and Spot the Patterns
L3 ApplyMaterials: Graph paper, pen.
Predict: What is the sum of the entries of row \(n\)? What about the alternating sum?
- Write rows 0 through 8 of Pascal's Triangle.
- Sum the entries of each row. Pattern? (Hint: \(2^n\).)
- Alternating sum (\(+,-,+,-,\ldots\)): always 0 for \(n\ge 1\). Why?
- Read along a diagonal: 1, 2, 3, 4, 5, ... (natural numbers); next diagonal: 1, 3, 6, 10, 15, ... (triangular numbers).
- Sum down each "shallow diagonal": 1, 1, 2, 3, 5, 8, ... — the Fibonacci numbers!
Row sum = \(2^n\) because setting \(a=b=1\) in the theorem gives \(\sum_k\binom{n}{k}=2^n\). Alternating sum = \((1-1)^n=0\) for \(n\ge 1\). The natural-numbers and triangular-numbers diagonals correspond to \(\binom{n}{1}\) and \(\binom{n}{2}\). The Fibonacci-shallow-diagonal identity \(F_n=\sum_{k}\binom{n-k-1}{k}\) is one of the most beautiful consequences of the triangle's structure.
Competency-Based Questions
Scenario: A research team analysing genome sequences needs to count subsets of size \(k\) from \(n\) markers. Each row of Pascal's Triangle gives the count for fixed \(n\) across all \(k\), so the team uses the triangle as a fast lookup.
Q1. How many terms appear in the expansion of \((x+y)^{20}\)?
L1 RememberAnswer: (c) 21. \(n+1=21\) terms.
Q2. The coefficient of \(x^3\) in \((1+x)^7\) is:
L3 ApplyAnswer: \(\binom{7}{3}=\dfrac{7\cdot 6\cdot 5}{3!}=35\).
Q3. (Fill the blank) The sum of the entries in row 10 of Pascal's Triangle is ____, because setting \(a=b=1\) in the binomial theorem gives ____.
L2 UnderstandAnswer: \(2^{10}=1024\). Setting \(a=b=1\) gives \((1+1)^n=\sum_k\binom{n}{k}\).
Q4. (T/F) "Pascal's identity \(\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}\) is the only ingredient needed (besides base case) for the inductive proof of the binomial theorem." Justify.
L5 EvaluateTrue. The induction step multiplies \((a+b)^m\) by \((a+b)\), splits into two sums shifted by one, and combines using Pascal. No other identity is needed. The base case \(n=1\) is trivial. So Pascal + base case → entire theorem.
Q5. Design: estimate \((1.01)^{1000}\) using only the first 4 terms of the binomial expansion. Compare with the actual value (≈ 21,000).
L6 CreateSolution: \((1+0.01)^{1000}\approx \binom{1000}{0}+\binom{1000}{1}(0.01)+\binom{1000}{2}(0.01)^2+\binom{1000}{3}(0.01)^3=1+10+49.95+166.17\approx 227\). The first 4 terms underestimate badly because the series converges slowly here. The true value is ≈ 20,959 — many more terms are needed when the exponent is large. (Calculus tools — Taylor series — accelerate this.)
Assertion–Reason Questions
Assertion (A): The number of terms in \((a+b)^{15}\) is 16.
Reason (R): The binomial expansion of \((a+b)^n\) has \(n+1\) terms, indexed \(k=0,1,\ldots,n\).
Reason (R): The binomial expansion of \((a+b)^n\) has \(n+1\) terms, indexed \(k=0,1,\ldots,n\).
Answer: (a). R is the rule that gives A directly.
Assertion (A): Each interior entry of Pascal's Triangle is the sum of the two entries directly above it.
Reason (R): \(\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}\) (Pascal's identity).
Reason (R): \(\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}\) (Pascal's identity).
Answer: (a). The two entries above \(\binom{n}{k}\) in the triangle are \(\binom{n-1}{k-1}\) and \(\binom{n-1}{k}\), and R is exactly the equation that explains A.
Assertion (A): \(\sum_{k=0}^{n}\binom{n}{k}=2^n\).
Reason (R): Setting \(a=b=1\) in the binomial theorem gives \((1+1)^n=\sum_k\binom{n}{k}\).
Reason (R): Setting \(a=b=1\) in the binomial theorem gives \((1+1)^n=\sum_k\binom{n}{k}\).
Answer: (a). R is the substitution that produces A.
Frequently Asked Questions
What is the Binomial Theorem?
For any positive integer n, (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + ⋯ + ⁿCₙ bⁿ. The coefficients ⁿCₖ are called binomial coefficients and equal n! / (k!(n−k)!).
What is Pascal's Triangle?
A triangular array where each entry is the sum of the two entries directly above it. The kth entry of row n is the binomial coefficient ⁿCₖ. Row 0 starts with 1; row 1 is 1 1; row 2 is 1 2 1; row 3 is 1 3 3 1; and so on.
How is the Binomial Theorem proved?
By induction on n. Base case n = 1 is trivial. Assuming the theorem for n, multiply both sides by (a + b) and use Pascal's identity ⁿCₖ + ⁿCₖ₋₁ = ⁿ⁺¹Cₖ to collect terms — yielding the theorem for n + 1.
How many terms are in (a + b)ⁿ?
Exactly n + 1 terms, indexed by k = 0, 1, 2, …, n. The kth term has the form ⁿCₖ aⁿ⁻ᵏ bᵏ.
Why are powers of a decreasing while powers of b increasing?
Each term is a product where one of {a, b} is chosen at each of the n binomial factors. If b is chosen k times, a is chosen the remaining n − k times — giving aⁿ⁻ᵏ bᵏ. The number of ways to choose k positions out of n is ⁿCₖ, the coefficient.
Who first wrote down Pascal's Triangle?
The triangle pattern appeared in Indian, Persian, Chinese, and European sources before Pascal. The Indian scholar Pingala (3rd–2nd century BCE) described what is now called the Meru Prastara, equivalent to the triangle. Pascal published a comprehensive Traité du Triangle Arithmétique posthumously in 1665, after which the triangle bore his name.
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