Binomial Theorem
Pascal's triangle patterns, binomial expansion formula (a+b)ⁿ & general term applications.
Topic 7.1 – 7.2: Pascal's Triangle & Symmetry of Binomial Coefficients
Imagine stacking crispy banana chips into a triangular pyramid! Place
1 at the tip, keep 1 along both slanted outer edges, and fill every inside spot by adding the TWO numbers right above it! Row 4 turns into 1, 4, 6, 4, 1 — amazingly, those exact pyramid numbers are the magic coefficients of (a + b)⁴!TRIANGLE RULEEach inner entry = sum of 2 numbers above:
ⁿCᵣ = ⁿ⁻¹Cᵣ₋₁ + ⁿ⁻¹CᵣTERM COUNT & SYMMETRYPower
n has n + 1 terms; mirror twins ⁿCᵣ = ⁿCₙ₋ᵣROW SUM MAGICSum of Row
n entries = ⁿC₀ + ⁿC₁ + ... + ⁿCₙ = 2ⁿ!🧠 Memory Trick: Row
n starts and ends with 1, reads the same forwards & backwards (palindrome!), and ALWAYS adds up to 2ⁿ!💡 Simple Kid-Check: Look at
(a + b)² = 1a² + 2ab + 1b² (Row 2: 1, 2, 1) and (a + b)³ = 1a³ + 3a²b + 3ab² + 1b³ (Row 3: 1, 3, 3, 1). You don't have to multiply brackets manually — just read Row n!| Power n | Pascal Row Coefficients (ⁿC₀ ... ⁿCₙ) | Terms (n+1) | Row Sum (2ⁿ) |
|---|---|---|---|
| n = 0 | 1 | 1 | 1 |
| n = 1 | 1 1 | 2 | 2 |
| n = 2 | 1 2 1 | 3 | 4 |
| n = 3 | 1 3 3 1 | 4 | 8 |
| n = 4 | 1 4 6 4 1 | 5 | 16 |
| n = 5 | 1 5 10 10 5 1 | 6 | 32 |
Watch TWO sneaky 1-mark traps: (1) How many terms does
(a + b)ⁿ have? Answer is n + 1 (NOT n!). (2) If a trinomial hides a binomial cube like (1 + 3x + 3x² + x³)⁴ = ((1 + x)³)⁴ = (1 + x)¹², the total number of distinct terms is 12 + 1 = 13, NOT 4 + 1 = 5!Topic 7.2: Binomial Expansion Formula for Positive Integer n
What if someone asks you to expand
(2x + 3y)⁶? Drawing six rows of multiplication would fill two pages! Instead, the Binomial Theorem gives a rhythm like two kids on an escalator: a walks DOWN from power n to 0, while b walks UP from power 0 to n, and Pascal's coefficients ⁿCᵣ cheer them on!BINOMIAL FORMULA
(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ... + ⁿCₙbⁿPOWER BALANCEIn EVERY term
ⁿCᵣ aⁿ⁻ʳ bʳ, exponents satisfy (n − r) + r = n!INDEX RULE
r goes 0, 1, ..., n (b's power = r, matching bottom of ⁿCᵣ!)🧠 Memory Trick: Check your expansion in 2 seconds: (1)
n + 1 terms? (2) First exponent + Second exponent = n in EVERY term? (3) Coefficients mirror?💡 Simple Kid-Check: Think of expanding
(a + b)ⁿ as choosing r copies of b and (n − r) copies of a from n identical brackets — there are ⁿCᵣ ways to pick those r brackets!
Master Formula:
(a + b)ⁿ = ∑r=0n ⁿCᵣ · an−r · br (for any positive integer n)
🧒 Step-by-Step Expansion of
Identify
(x² + 3x)⁴:Identify
a = x², b = 3/x, n = 4 (Pascal row: 1, 4, 6, 4, 1):= 1(x²)⁴ + 4(x²)³(3x) + 6(x²)²(3x)² + 4(x²)(3x)³ + 1(3x)⁴= x⁸ + 12x⁵ + 54x² + 108x + 81x⁴!
When expanding
(2x − 3y)⁵ or (x/3 + 1/x)⁵, students forget to apply the exponent to the numerical coefficient inside the bracket! Write parentheses ALWAYS: (2x)³ = 8x³ (NOT 2x³!), and keep the minus sign INSIDE b = (−3y) so signs alternate cleanly!Topic 7.2: Special Cases: (1 + x)ⁿ, (1 − x)ⁿ & Sign Alternation
Set
a = 1 and suddenly every power of 1 melts into plain 1 — leaving the sleekest expansion in algebra: (1 + x)ⁿ! Swap +x to −x and watch the signs blink + − + − like Onam fairy lights! Even cooler: plug in x = 1 and you prove 2ⁿ = sum of Pascal row in one line!(1 + x)ⁿ & (1 − x)ⁿ
(1±x)ⁿ = ⁿC₀ ± ⁿC₁x + ⁿC₂x² ± ... + (±1)ⁿⁿCₙxⁿALL COEFFICIENTSPut
x = 1 in (1+x)ⁿ → ⁿC₀ + ⁿC₁ + ... + ⁿCₙ = 2ⁿEVEN = ODD SUMPut
x = −1 → ⁿC₀ + ⁿC₂ + ... = ⁿC₁ + ⁿC₃ + ... = 2ⁿ⁻¹🧠 Memory Trick: In
(1 − x)ⁿ, look at the bottom index r of ⁿCᵣ: EVEN r → PLUS (+), ODD r → MINUS (−)!💡 Simple Kid-Check: Why does
(1 − x)ⁿ alternate signs? Because (−x)ʳ = (−1)ʳ xʳ! Whenever r is even (0, 2, 4), minus pairs cancel into +; whenever r is odd (1, 3, 5), one minus survives!
Number-Crunching Power Trick: Evaluate
Write
(99)⁵ or (102)⁵ without a calculator!Write
99 = 100 − 1 = 10² − 1 or 102 = 100 + 2 and expand!
🧒 Quick Numerical Example: Evaluate
(99)⁴ using Binomial Theorem.(100 − 1)⁴ = 100⁴ − 4(100)³(1) + 6(100)²(1)² − 4(100)(1)³ + 1⁴= 100,000,000 − 4,000,000 + 60,000 − 400 + 1 = 96,059,601!
When computing
(98)⁵ = (100 − 2)⁵, students often lose marks by flipping the alternating signs incorrectly OR dropping a zero when writing 100⁴ = 10⁸ (8 zeros, NOT 6!). Double-check the zero count on every power of 100 (100^k has 2k zeros!) and verify Even r = +, Odd r = −!Topic 7.2: Simplifying Surds & Comparing (1.01)¹⁰⁰⁰⁰⁰⁰ vs 10000
Which is bigger:
(1.01)¹⁰⁰⁰⁰⁰⁰ or 10,000? Don't guess — slice 1.01 = 1 + 0.01! Just the FIRST TWO positive binomial terms give 1 + 1,000,000 × 0.01 = 10,001 — already beating 10,000 with a million positive terms still left! And when you add (a + √b)ⁿ + (a − √b)ⁿ, every messy square root magically cancels!SUM (a+b)ⁿ + (a−b)ⁿ
2[ⁿC₀aⁿ + ⁿC₂aⁿ⁻²b² + ...] (ONLY EVEN r survive!)DIFF (a+b)ⁿ − (a−b)ⁿ
2[ⁿC₁aⁿ⁻¹b + ⁿC₃aⁿ⁻³b³ + ...] (ONLY ODD r survive!)GIANT POWER BOUNDFor
d > 0: (1 + d)ⁿ = 1 + nd + [positive terms] > 1 + nd!🧠 Memory Trick: + keeps EVEN terms doubled (so square roots of
b vanish because b², b⁴ become integers!); − keeps ODD terms doubled!💡 Simple Kid-Check: Adding
(a + b)⁴ + (a − b)⁴ kills every term with an odd power of b (+ meets −), while doubling every even-power term! That's why (√3 + √2)⁴ + (√3 − √2)⁴ turns into a pure whole number with ZERO roots left!
🧒 Classic 3-Mark Surd Problem: Find
• Keep ONLY odd
• Plug
(a + b)⁴ − (a − b)⁴, hence evaluate (√3 + √2)⁴ − (√3 − √2)⁴:• Keep ONLY odd
r = 1, 3 doubled: 2[⁴C₁ a³b + ⁴C₃ a b³] = 2[4a³b + 4ab³] = 8ab(a² + b²).• Plug
a = √3, b = √2: 8(√3)(√2)(3 + 2) = 8√6(5) = 40√6!
🚀 NCERT Mega-Power Comparison: Compare
(1.01)¹⁰⁰⁰⁰⁰⁰ and 10,000:(1 + 0.01)¹⁰⁰⁰⁰⁰⁰ = ¹⁰⁰⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰⁰⁰C₁(0.01) + other positive terms= 1 + 10,000 + positive terms = 10,001 + positive terms > 10,000!
Watch the sign mistake in
(a + b)ⁿ − (a − b)ⁿ! The minus sign outside flips EVERY term of (a − b)ⁿ: so even terms (+ − +) become − + − and CANCEL OUT, while odd terms (− inside (a − b)ⁿ) become + and DOUBLE! Always remember: Plus keeps Even, Minus keeps Odd!Topic 7.2: Proving Divisibility Properties Using Binomial Expansion
How can you prove that
9ⁿ⁺¹ − 8n − 9 is divisible by 64 for EVERY positive integer n = 1, 2, 3, ..., infinity? You can't test infinitely many numbers — instead, crack open 9 as (1 + 8)! Every higher term of (1 + 8)ⁿ⁺¹ carries 8² = 64 inside it like a secret badge!SECRET SPLITRewrite base as
(1 + k) where k² (or k) divides the target!KEY IDENTITY
(1 + k)ᵐ = 1 + mk + ᵐC₂k² + ᵐC₃k³ + ...FACTOR STEPCancel
1 + mk with linear tail, then pull out k² × integer!🧠 Memory Trick: Spot the square root of the divisor! Divisor
64 → root is 8 → split 9 = 1 + 8! Divisor 49 → root is 7 → rewrite 2³ⁿ = 8ⁿ = (1 + 7)ⁿ!💡 Simple Kid-Check: Watch the trick in
3²ⁿ⁺² − 8n − 9: first rewrite 3²ⁿ⁺² = (3²)ⁿ⁺¹ = 9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹! Now (1 + 8) gives 8² = 64 cleanly!- 1Match Base to 1 + k: Here
k = 8because8² = 64. Write9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹. - 2Expand First Few Terms:
(1 + 8)ⁿ⁺¹ = 1 + (n+1)(8) + ⁿ⁺¹C₂(8²) + ⁿ⁺¹C₃(8³) + .... - 3Cancel Tail & Factor 64:
9ⁿ⁺¹ = (8n + 9) + 64 [ⁿ⁺¹C₂ + 8·ⁿ⁺¹C₃ + ...]⇒9ⁿ⁺¹ − 8n − 9 = 64m(wherem ∈ ℤ)!
🧪 Try Another Classic: Prove
Split
4ⁿ − 3n − 1 is divisible by 9.Split
4 = 1 + 3 → (1 + 3)ⁿ = 1 + 3n + ⁿC₂·3² + ... = (3n + 1) + 9·(integer) ⇒ 4ⁿ − 3n − 1 = 9k!
Watch TWO classic proof errors: (1) Starting directly with
(9)ⁿ⁺¹ = (8 + 1)ⁿ⁺¹ BACKWARDS (putting 8 first instead of 1 first)! Always put 1 first — (1 + 8)ⁿ⁺¹ — so the linear part sits neatly at the front (1 + 8(n+1))! (2) Leaving the remaining bracket undefined — clearly state where m is an integer to earn full 3/4 marks!Topic 7.2: Finding Specific Terms, Constant Terms & Coefficient Comparisons
Need ONLY the constant term (the term with no
x) inside (x² + 1/x)⁹? Don't waste 10 minutes expanding all 10 terms! Fire the Binomial sniper scope — the General Term Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ — calculate the net exponent of x in terms of r, set it equal to target power (0!), and grab the exact answer in 3 lines!GENERAL TERM
Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ʳ · bʳ (Always r + 1 on left, r on right!)COEFFICIENT TARGETCollect total power of
x → equate to desired power k (0 for constant term!)MIDDLE TERM(S)
n even → 1 middle: T(n/2)+1 | n odd → 2 middles!🧠 Memory Trick: Want the 4th term
T₄? Write T3+1 so r = 3 (ALWAYS subtract 1 from the term number)!💡 Simple Kid-Check: Why is the
(r + 1)th term written Tᵣ₊₁? Because the very FIRST term T₁ starts at r = 0! So T₁ = T₀₊₁, T₂ = T₁₊₁, and in general Tₖ uses r = k − 1!
🎯 Sniper Example 1 (Constant Term): Find the term independent of
For term independent of
So the constant term is
x in (x² + 1x)⁶:Tᵣ₊₁ = ⁶Cᵣ · (x²)⁶⁻ʳ · (x⁻¹)ʳ = ⁶Cᵣ · x¹²⁻²ʳ⁻ʳ = ⁶Cᵣ · x¹²⁻³ʳ.For term independent of
x (x⁰): set 12 − 3r = 0 ⇒ r = 4.So the constant term is
T₅ = ⁶C₄ = ⁶C₂ = 15!
⚖️ Sniper Example 2 (Equal Terms): Find
a if T₁₇ = T₁₈ in (2 + a)⁵⁰:⁵⁰C₁₆ · 2³⁴ · a¹⁶ = ⁵⁰C₁₇ · 2³³ · a¹⁷ → divide ⁵⁰C₁₆ / ⁵⁰C₁₇ = 17/34 = 1/2 → 2 · (1/2) = a ⇒ a = 1!
Watch TWO lethal exam traps: (1) Finding the
5th term using r = 5 instead of r = 4 (T₅ = T₄₊₁)! (2) When finding a coefficient like x⁵ in (x + 3)⁸, if r turns out to be a fraction or negative or > n, that power of x simply DOES NOT EXIST (coefficient = 0)!Solved Previous Year Questions (Kerala DHSE & NCERT)
Q1 (Kerala HSE March • 1 Marks): Find the number of terms in the expansion of (1 + 3x + 3x² + x³)⁴.
Notice
So
Number of terms =
1 + 3x + 3x² + x³ = (1 + x)³.So
((1 + x)³)⁴ = (1 + x)¹².Number of terms =
12 + 1 = 13.Q2 (Kerala HSE Say • 1 Marks): Evaluate ⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅.
Put
x = 1 in (1 + x)⁵: sum = 2⁵ = 32.Q3 (Kerala HSE March • 2 Marks): Expand (x/3 + 1/x)⁵ using Binomial Theorem.
⁵C₀(x/3)⁵ + ⁵C₁(x/3)⁴(1/x) + ⁵C₂(x/3)³(1/x)² + ⁵C₃(x/3)²(1/x)³ + ⁵C₄(x/3)(1/x)⁴ + ⁵C₅(1/x)⁵= x⁵243 + 5x³81 + 10x27 + 109x + 53x³ + 1x⁵.Q4 (NCERT / DHSE Model • 2 Marks): Using Binomial Theorem, evaluate (98)⁵.
Write
98 = 100 − 2:(100 − 2)⁵ = 100⁵ − 5(100⁴)(2) + 10(100³)(4) − 10(100²)(8) + 5(100)(16) − 32= 10,000,000,000 − 1,000,000,000 + 40,000,000 − 800,000 + 8,000 − 32 = 9,039,207,968.Q5 (Kerala HSE March • 3 Marks): Find (a + b)⁴ − (a − b)⁴. Hence evaluate (√3 + √2)⁴ − (√3 − √2)⁴.
Subtracting cancels even terms and doubles odd terms:
Setting
(a + b)⁴ − (a − b)⁴ = 2[⁴C₁a³b + ⁴C₃ab³] = 8ab(a² + b²).Setting
a = √3, b = √2: 8(√3)(√2)(3 + 2) = 40√6.Q6 (Kerala HSE March • 3 Marks): Show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64, whenever n is a positive integer.
Write
9 = 1 + 8:9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ = 1 + (n+1)(8) + ⁿ⁺¹C₂·8² + ⁿ⁺¹C₃·8³ + ...9ⁿ⁺¹ = (8n + 9) + 64[ⁿ⁺¹C₂ + 8·ⁿ⁺¹C₃ + ...]⇒ 9ⁿ⁺¹ − 8n − 9 = 64m (where m is a positive integer). Hence divisible by 64!Q7 (Kerala HSE Say • 3 Marks): Find the coefficient of x⁵ in (x + 3)⁸.
Tᵣ₊₁ = ⁸Cᵣ · x⁸⁻ʳ · 3ʳ.Set
8 − r = 5 ⇒ r = 3.Coefficient =
⁸C₃ · 3³ = 56 × 27 = 1,512.Q8 (NCERT / Kerala HSE • 4 Marks): Find a if the 17th and 18th terms of the expansion (2 + a)⁵⁰ are equal.
T₁₇ = ⁵⁰C₁₆ · 2³⁴ · a¹⁶ and T₁₈ = ⁵⁰C₁₇ · 2³³ · a¹⁷.Setting
T₁₇ = T₁₈:a = ⁵⁰C₁₆ · 2³⁴⁵⁰C₁₇ · 2³³ = 2 × 1734 = 1.