Binomial Theorem

2.1 Core concepts

• The Binomial Theorem here applies to positive integral indices only; the general case for integral or rational n is out of scope (NCERT §7.1, p. 126). General binomial series for non-integer exponents involve infinite series and convergence considerations.
• Motivation: numerical computation of high powers such as (98)^5 or (101)^6 by repeated multiplication is laborious, so a single formulaic expansion is sought (NCERT §7.1, p. 126).
• The expansions of (a + b)^0, (a + b)^1, (a + b)^2, (a + b)^3, (a + b)^4 are tabulated and three structural observations are extracted: (i) number of terms is one more than the index, (ii) the power of a decreases by 1 and the power of b increases by 1 in successive terms, (iii) the sum of indices of a and b in every term equals the index of (a + b) (NCERT §7.2, p. 126).
• The coefficients are arranged in a triangular array — 1 at the apex, with each interior entry obtained by adding the two entries immediately above it — known as Pascal's triangle, also called Meru Prastara after Pingla (NCERT §7.2, pp. 127–128). The triangle predates Pascal; Indian mathematician Pingla described it in his Chandahsastra (~200 BCE).
• For ready use, Pascal's triangle is rewritten with entries as combinations nCr = n! / [r!(n – r)!], 0 ≤ r ≤ n, with nC0 = 1 = nCn; this lets us write the row for any index without writing all preceding rows (NCERT §7.2, p. 128).
• For index 7, the row is 7C0, 7C1, 7C2, 7C3, 7C4, 7C5, 7C6, 7C7, giving (a + b)^7 = 7C0 a^7 + 7C1 a^6 b + 7C2 a^5 b^2 + 7C3 a^4 b^3 + 7C4 a^3 b^4 + 7C5 a^2 b^5 + 7C6 a b^6 + 7C7 b^7 (NCERT §7.2, p. 128).
• Binomial Theorem statement (§7.2.1): For any positive integer n, (a + b)^n = nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2 + … + nC(n-1) a b^(n-1) + nCn b^n, equivalently (a + b)^n = Σ_{k=0}^{n} nCk a^(n-k) b^k (NCERT §7.2.1, pp. 129–130).
• Proof is by the principle of mathematical induction: base step at n = 1; inductive step uses kCr + kC(r-1) = (k+1)Cr together with kC0 = (k+1)C0 = 1 and kCk = (k+1)C(k+1) = 1 to pass from P(k) to P(k+1) (NCERT §7.2.1, p. 129). Pascal's identity is the algebraic engine of the inductive step.
• Worked illustration: (x + 2)^6 = x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64 (NCERT §7.2.1, p. 129). Note that powers of 2 grow as 1, 2, 4, 8, 16, 32, 64 while binomial coefficients are 1, 6, 15, 20, 15, 6, 1.
• Five Observations on the expansion (§7.2.1): sigma form, the term nCr is called a binomial coefficient, there are (n + 1) terms, the index of a runs n, n–1, …, 0 while the index of b runs 0, 1, …, n, and the sum of the indices of a and b in every term equals n (NCERT §7.2.1, p. 130).
• Special case 1 — (x – y)^n: Setting a = x, b = –y gives (x – y)^n = nC0 x^n – nC1 x^(n-1) y + nC2 x^(n-2) y^2 – … + (–1)^n nCn y^n; the signs alternate. Illustration: (x – 2y)^5 = x^5 – 10x^4 y + 40x^3 y^2 – 80x^2 y^3 + 80 x y^4 – 32 y^5 (NCERT §7.2.2, p. 130).
• Special case 2 — (1 + x)^n: Setting a = 1, b = x gives (1 + x)^n = nC0 + nC1 x + nC2 x^2 + … + nCn x^n. In particular, putting x = 1 yields 2^n = nC0 + nC1 + nC2 + … + nCn (NCERT §7.2.2, pp. 130–131).
• Special case 3 — (1 – x)^n: Setting a = 1, b = –x gives (1 – x)^n = nC0 – nC1 x + nC2 x^2 – … + (–1)^n nCn x^n. Putting x = 1 yields 0 = nC0 – nC1 + nC2 – … + (–1)^n nCn (NCERT §7.2.2, p. 131).
• Numerical applications (§7.2.2 examples): (98)^5 is computed as (100 – 2)^5 = 9 039 207 968; (1.01)^1000000 > 10 000 by retaining the first two non-negative binomial terms 1 + 1 000 000 × 0.01 (NCERT pp. 131–132).
• Divisibility application (Example 4): Using (1 + 5)^n = 1 + 5n + 5^2·nC2 + 5^3·nC3 + … + 5^n, one obtains 6^n – 5n = 25k + 1 where k = nC2 + 5·nC3 + … + 5^(n-2), proving that 6^n – 5n leaves remainder 1 on division by 25 (NCERT §7.2.2, p. 132).
• The triangular array of binomial coefficients is called Pascal's triangle (NCERT Summary, p. 133).
• The general term of (a + b)^n — the (r + 1)-th term — is T_{r+1} = nCr a^(n-r) b^r. This single formula handles every "find coefficient of x^k" and "find middle term" problem.
• Two identities from Permutations and Combinations (Ch. 6) are useful here: symmetry nCr = nC(n−r) and Pascal's identity nCr + nC(r−1) = (n+1)Cr. These let you compute large nCr by reducing to small r quickly.
• "Middle term" considerations: for the binomial (a + b)^n, if n is even there is one middle term at position (n/2) + 1; if n is odd there are two middle terms at positions (n+1)/2 and (n+3)/2.
• The ratio of consecutive terms is T_{r+2}/T_{r+1} = [(n − r)/(r + 1)] · (b/a). This ratio is monotonic and helps locate the largest term (greatest binomial coefficient times power).
• The greatest binomial coefficient: in row n of Pascal's triangle, the largest entry is nC⌊n/2⌋ (one or two values for even/odd n). For example, the largest binomial coefficient in (a + b)^10 is 10C5 = 252.
• Connection to combinatorics: nCk counts the number of k-element subsets of an n-element set; the binomial theorem can be read as a generating-function identity that encodes all of these subset counts simultaneously.
• The numerical approximation philosophy: for (1 + x)^n with |x| small, the first two terms 1 + nx are an excellent approximation; this is the leading-order Taylor approximation, formally justified in Class XII analysis but already useful here.
• A useful identity derived from sub-cases of the binomial theorem: Σ k·nCk = n·2^(n−1) (obtained by differentiating (1+x)^n and setting x = 1). This and similar identities show how calculus tools enrich combinatorial results.
• Connection to probability: the binomial probability distribution P(X = k) = nCk p^k (1−p)^(n−k) has its mass function built directly from the binomial theorem with a = p, b = 1−p, giving Σ_k P(X = k) = (p + 1 − p)^n = 1.
• For a quick "row of Pascal's triangle" without computing all factorials: start with 1 at both ends; compute each interior entry as the sum of the two above. This builds up row by row efficiently for small n (≤ 15).
• The theorem extends conceptually but not computationally to (a + b + c)^n via the multinomial theorem; this is beyond Class XI but appears occasionally in JEE-style questions.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 7.1 (p. 127): Tabular display of coefficients for indices 0 through 4 — the raw form from which Pascal's pattern is read off.
• Fig 7.2 (p. 127): Pascal's triangle with explicit arrows/triangular markers showing how adjacent entries in row n add to give the next entry in row n+1.
• Fig 7.3 (p. 128): Pascal's triangle rewritten with combination notation kCr (each cell shows nCr and the numerical value), exhibited up to row n = 5.
• Portrait of Blaise Pascal (1623–1662) on p. 126 alongside the introduction, linking the topic to its historical figure.
• Process for expanding using Pascal's triangle: pick the row matching the index, attach decreasing powers of the first quantity and increasing powers of the second (worked example on p. 127 for (2x + 3y)^5).
• Process — find (r+1)-th term: identify n and the binomial (a+b)^n; substitute into T_{r+1} = nCr a^(n-r) b^r; simplify.
• Process — find coefficient of x^k in (a + bx)^n: set up T_{r+1}, identify which r makes the x-exponent equal to k, solve for r, plug back.
• Process — middle term: if n even, middle term is the (n/2 + 1)-th; if n odd, there are two middle terms at positions (n+1)/2 and (n+3)/2.
• Process — numerical approximation: write the number as (a + b)^n where b is small; truncate the expansion after a few terms.

2.4 Common confusions / NTA trap points

• Number of terms vs. index. The expansion of (a + b)^n has n + 1 terms, not n. The first-term index of a is n, not n – 1 (NCERT §7.2.1 obs. 3–4, p. 130).
• Sign pattern in (x – y)^n and (1 – x)^n. Coefficients of even powers of the negative quantity stay positive, odd powers turn negative; students often misplace a sign. The r-th coefficient carries (–1)^r, not (–1)^(n-r) (NCERT §7.2.2, pp. 130–131).
• Reading off the right term. In (a + b)^n = Σ nCk a^(n-k) b^k, the term with b^k is the (k + 1)-th term, not the k-th — because the indexing starts at k = 0. A frequent NTA distractor is to swap nCk a^(n-k) b^k with nCk a^k b^(n-k).
• Sum-of-indices invariance. Every term in the expansion has indices summing to n. An option in which the indices sum to n – 1 or n + 1 is automatically a distractor (NCERT §7.2.1 obs. 5, p. 130).
• Pascal's triangle row vs. index. Row labelled by index n contains n + 1 entries; the first entry is nC0 = 1, not nC1. Misreading the row offset produces wrong coefficients (NCERT Fig 7.3, p. 128).
• (1.01)^1000000 style comparison. Keep only the first one or two terms of (1 + 0.01)^1000000 and discard the rest because they are positive — discarding a negative term would be illegitimate (NCERT Example 3, p. 132).
• Middle term count. For n odd, there are two middle terms, not one. CUET sometimes traps with a one-term option when n is odd.
• Wrong substitution for x in (1 + x)^n. To get the sum 2^n, substitute x = 1; substituting x = 2 gives 3^n, not 2^n.
• Treating coefficient and term as the same. "Coefficient of x^k" is the number multiplying x^k; the "term containing x^k" includes the x^k.
• Forgetting the factor (b)^k in coefficient hunts. When the binomial is (a + bx)^n with b ≠ 1, the coefficient of x^k is nCk·a^(n-k)·b^k.
• Mis-applying the theorem to non-integer n. The Class XI version is only for positive integer n; negative or fractional n is the general binomial series (not in syllabus).
• Confusing nCr with nPr. Combinations have an r! in the denominator; permutations do not.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT §7.2.1 worked, p. 129). Expand (x + 2)^6.

Step 1 — coefficients (Pascal's row 6): 1, 6, 15, 20, 15, 6, 1. Step 2 — assemble terms: T_1 = x^6, T_2 = 6x^5·2, T_3 = 15x^4·4, … Step 3 — simplify: x^6 + 12x^5 + 60x^4 + 160x^3 + 240x^2 + 192x + 64. Answer: as above.

Example B (NCERT §7.2.2 worked, p. 130). Expand (x − 2y)^5.

Step 1 — apply (x + (−2y))^5: coefficients (1, 5, 10, 10, 5, 1). Step 2 — alternating signs from −2y: signs (+, −, +, −, +, −). Step 3 — simplify: x^5 − 10x^4 y + 40x^3 y^2 − 80x^2 y^3 + 80xy^4 − 32y^5. Answer: as above.

Example C (NCERT §7.2.2, p. 131). Compute (98)^5.

Step 1 — write 98 = 100 − 2: apply (100 − 2)^5. Step 2 — expand: 100^5 − 5·100^4·2 + 10·100^3·4 − 10·100^2·8 + 5·100·16 − 32. Step 3 — sum: 10000000000 − 1000000000 + 40000000 − 800000 + 8000 − 32 = 9039207968.

Example D (NCERT Example 4, p. 132). Show 6^n − 5n leaves remainder 1 on division by 25.

Step 1 — expand (1 + 5)^n: 1 + 5n + 25·nC2 + 125·nC3 + … + 5^n. Step 2 — subtract 5n + 1: 6^n − 5n − 1 = 25(nC2 + 5·nC3 + … + 5^(n−2)). Step 3 — conclude: RHS divisible by 25 ⇒ 6^n − 5n ≡ 1 (mod 25). QED.

Example E (Finding a specific term). Find the coefficient of x^7 in the expansion of (1 + x)^10.

Step 1 — apply (1+x)^n formula: coefficient of x^k is nCk. Step 2 — substitute n = 10, k = 7: 10C7. Step 3 — evaluate: 10C7 = 10C3 = 120. Answer: 120.