Binomial Theorem 🔺

(a+b)¹⁰⁰ expand karna? Haath se impossible hai — lekin Binomial Theorem se seconds mein! Pascal ka 400 saal purana discovery.

🔢 Pascal ka Triangle

Blaise Pascal (1623–1662) ek French mathematician the. Unhone observe kiya ki (a+b)ⁿ expand karne par jo coefficients aate hain woh ek beautiful triangle pattern follow karte hain. Har number uske upar wale dono numbers ka sum hota hai. Yeh pattern India mein pehle hi Halayudha (10th century) ne discover kiya tha — "Meru Prastara" ke naam se! Pascal ka naam Western world mein popular hua.

1. Pascal's Triangle — Coefficients ka Pattern

Neeche Pascal's Triangle dekho — (a+b)⁰ se (a+b)⁵ tak ke coefficients:

n=0 se n=5 (a+b)ⁿ ke coefficients:

n=0:

1

n=1:

1
1

n=2:

1
2
1

n=3:

1
3
3
1

n=4:

1
4
6
4
1

n=5:

1
5
10
10
5
1

Rule: Har cell = uske upar wale dono cells ka sum | Har row ke corner mein 1 hamesha!

Yeh coefficients exactly ⁿC₀, ⁿC₁, ⁿC₂, ... ⁿCₙ hain — yahi connection Combinations aur Binomial Theorem ko jodta hai!

2. Binomial Theorem — Main Formula

(a+b)ⁿ = Σ(r=0 to n) ⁿCᵣ · aⁿ⁻ʳ · bʳ

= ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ⁿC₃aⁿ⁻³b³ + ... + ⁿCₙbⁿ

Kaise yaad karo: Har term mein a aur b ke powers ka sum n hota hai (aⁿ⁻ʳ × bʳ → powers = n−r+r = n). r = 0 se shuru hota hai n tak. Coefficient = ⁿCᵣ (Combinations!)

✅ (x+y)⁴ expand karo

n=4. Row 4 of Pascal: 1, 4, 6, 4, 1 (yani ⁴C₀,⁴C₁,⁴C₂,⁴C₃,⁴C₄)
(x+y)⁴ = 1·x⁴ + 4·x³y + 6·x²y² + 4·xy³ + 1·y⁴
= x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

Notice: x ka power ghatta hai (4,3,2,1,0), y ka power badhta hai (0,1,2,3,4)!

✅ (2x − 3y)³ expand karo

a = 2x, b = −3y, n = 3. Coefficients: 1, 3, 3, 1
= (2x)³ + 3(2x)²(−3y) + 3(2x)(−3y)² + (−3y)³
= 8x³ + 3·4x²·(−3y) + 3·2x·9y² + (−27y³)
= 8x³ − 36x²y + 54xy² − 27y³

Jab b negative ho, alternating signs aate hain: +, −, +, −

3. General Term Tᵣ₊₁

Kisi bhi ek specific term ko nikalna ho — direct formula hai. (r+1)th term ko T(r+1) likhte hain:

T(r+1) = ⁿCᵣ · aⁿ⁻ʳ · bʳ

r = 0 → T₁ (first term), r = 1 → T₂ (second term), aur aise hi aage. Ek specific term ka coefficient ya variable power nikalna hai toh pehle r dhundo!

✅ (x + 2)⁸ mein x⁵ wala term nikalo

T(r+1) = ⁸Cᵣ · x⁸⁻ʳ · 2ʳ
x ka power chahiye = 5 → 8−r = 5 → r = 3
T₄ = ⁸C₃ · x⁵ · 2³ = 56 · 8 · x⁵ = 448x⁵

Steps: pehle r set karo, phir coefficient calculate karo!

✅ (x/2 − 2/x)⁶ mein x² wala term nikalo

a = x/2, b = −2/x, n = 6
T(r+1) = ⁶Cᵣ · (x/2)⁶⁻ʳ · (−2/x)ʳ
= ⁶Cᵣ · x⁶⁻ʳ/2⁶⁻ʳ · (−2)ʳ/xʳ = ⁶Cᵣ · (−1)ʳ · 2ʳ/2⁶⁻ʳ · x⁶⁻ʳ⁻ʳ
x ka power = 6−2r = 2 → r = 2
T₃ = ⁶C₂ · (−1)² · 2²/2⁴ · x² = 15 · 1 · (4/16) · x² = 15/4 · x² = (15/4)x²

4. Middle Term

(a+b)ⁿ mein n+1 terms hote hain. Middle term:

5. Important Special Cases — Binomial Sums

a=b=1 rakho, ya a=1, b=−1 rakho → powerful results milte hain:

(1+1)ⁿ = 2ⁿ → ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ
(1−1)ⁿ = 0 → ⁿC₀ − ⁿC₁ + ⁿC₂ − ... = 0
Even terms ka sum = Odd terms ka sum = 2ⁿ⁻¹

6. Approximations using Binomial

✅ (1.02)⁵ approximately nikalo

(1.02)⁵ = (1 + 0.02)⁵
≈ 1 + 5(0.02) + 10(0.02)² + ... (higher powers negligible)
≈ 1 + 0.10 + 10(0.0004) = 1 + 0.10 + 0.004 = 1.104

Real value = 1.10408 — approximation kaafi close hai!

7. Practice Questions

Q1. (1+x)¹⁰ mein x³ ka coefficient nikalo.

Solution

T₄: ¹⁰C₃ = 120. Coefficient = 120

Q2. (x − 1/x)⁶ mein constant term nikalo (x⁰ term).

Solution

T(r+1) = ⁶Cᵣ x⁶⁻ʳ(−1/x)ʳ = ⁶Cᵣ(−1)ʳx⁶⁻²ʳ. Constant: 6−2r=0 → r=3. T₄ = ⁶C₃(−1)³ = 20×(−1) = −20

Q3. ⁿC₀+ⁿC₁+...+ⁿCₙ = 1024. n ki value kya hai?

Solution

2ⁿ = 1024 = 2¹⁰ → n = 10

Q4. (3+x)⁶ ka 4th term likho.

Solution

T₄ = ⁶C₃ · 3³ · x³ = 20 × 27 × x³ = 540x³

Q5. (√x − √y)⁸ ka middle term nikalo.

Solution

n=8 (even) → Middle term = T₅ = ⁸C₄(√x)⁴(−√y)⁴ = 70 · x² · y² = 70x²y²

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