Class 11 Maths · Chapter 7
Short answer:
Class 11 Maths Chapter 7 — Binomial Theorem sikhata hai ki koi bhi (a + b)n jaisi expression ko, chahe n kitna bhi bada ho, bina full multiplication kiye directly expand kaise karein. Is chapter ke sirf teen sections hain — 7.1 Introduction, 7.2 Binomial Theorem for Positive Integral Indices (Pascal's Triangle sahit), aur 7.3 General and Middle Terms — aur do exercises: Exercise 7.1 (14 questions) aur Miscellaneous Exercise (6 questions). Core formula: Tr+1 = nCr an−r br.
Chapter 7 poore Class 11 Algebra ka sabse "formula-heavy lekin pattern-based" chapter hai — ek baar general term ki key mil jaaye, to poora chapter unlock ho jaata hai. Class 10 mein aapne (a+b)2 aur (a+b)3 ki identities ratt li thi; yahaan NCERT wahi idea ek general n tak le jaata hai, taaki (a+b)7 ho ya (a+b)100, aap koi bhi power seedha expand kar sako — bina ek-ek karke multiply kiye.
Ye chapter class 11 maths ncert solutions series mein exam ke hisaab se kaafi "safe scoring" chapter maana jaata hai — question pattern predictable hai: expand karo, ek specific term nikaalo, ya proof-type sawaal jo pure binomial expansion pe based hote hain. Neeche har question ka full step-by-step solution CBSE marking-scheme style mein diya gaya hai, jis structure mein aap Chapter 1 Sets se lekar Chapter 3 Trigonometric Functions tak ke NCERT solutions dekhte ho — same format, taaki revision easy rahe.
Syllabus note (ncert.nic.in ke current rationalised textbook, session 2026-27, ke against verify kiya gaya): is chapter ka number 7 hai — purani (2023 se pehle wali) numbering mein ye Chapter 8 tha, kyunki "Principle of Mathematical Induction" aur "Mathematical Reasoning" dono chapters syllabus se hata diye gaye hain. Agar kahin purani numbering "Chapter 8: Binomial Theorem" dikhe to woh stale hai — current session ke liye Chapter 7 hi sahi hai.
Chapter 7 Summary — 5 Minute Revision
Binomial Theorem chapter teen ideas pe khada hai. Pehla — expansion: (a+b)n ka har term nCr an−rbr se milta hai, aur in coefficients ka pattern Pascal's Triangle banata hai. Dusra — general term: Tr+1 formula se aap poora expansion likhe bina koi bhi specific term (5th term, x6 ka coefficient, term independent of x) seedha nikaal sakte ho. Teesra — middle term: n even ho to ek middle term hota hai ((n/2+1)th), n odd ho to do middle terms hote hain. Exercise 7.1 mein expansion practice aur "using binomial theorem evaluate" type questions hain (jaise 995, 1014), jabki Miscellaneous Exercise mein zyada conceptual proof-type sawaal hain (AP coefficients, divisibility proofs, product-expansion coefficient). Poore chapter mein 20 exercise questions aur is spec mein 9 additional in-text style worked examples cover kiye gaye hain.
In-Text Questions — Solutions
In-text Example 1: Expand (x + 2)6 using the Binomial Theorem.
Binomial theorem se: (a+b)n = Σr=0n nCr an−r br. Yahan a = x, b = 2, n = 6.
&sup6;C₀=1, &sup6;C₁=6, &sup6;C₂=15, &sup6;C₃=20, &sup6;C₄=15, &sup6;C₅=6, &sup6;C₆=1
(x+2)&sup6; = x⁶ + 6x⁵(2) + 15x⁴(4) + 20x³(8) + 15x²(16) + 6x(32) + 64
= x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64
In-text Example 2: Expand (x/3 − 2/x)4.
a = x/3, b = −2/x, n = 4. Coefficients: 1,4,6,4,1.
T₁ = (x/3)⁴ = x⁴/81
T₂ = 4(x/3)³(−2/x) = −8x²/27
T₃ = 6(x/3)²(−2/x)² = 6(x²/9)(4/x²) = 8/3
T₄ = 4(x/3)(−2/x)³ = 4(x/3)(−8/x³) = −32/(3x²)
T₅ = (−2/x)⁴ = 16/x⁴
(x/3 − 2/x)⁴ = x⁴/81 − 8x²/27 + 8/3 − 32/(3x²) + 16/x⁴
In-text Example 3: Using the Binomial Theorem, evaluate (98)4.
98 = 100 − 2. Isliye (100−2)⁴ expand karo:
(100−2)⁴ = 100⁴ − 4(100³)(2) + 6(100²)(4) − 4(100)(8) + 16
= 100,000,000 − 8,000,000 + 240,000 − 3,200 + 16
= 92,236,816
In-text Example 4: Expand (√2 + 1)5 using the Binomial Theorem.
a = √2, b = 1, n = 5, coefficients 1,5,10,10,5,1.
(√2)⁵ = 4√2, (√2)⁴=4, (√2)³=2√2, (√2)²=2
Terms = 4√2 + 5(4) + 10(2√2) + 10(2) + 5(√2) + 1
Rational terms jodo: 20 + 20 + 1 = 41. Irrational terms jodo: 4√2 + 20√2 + 5√2 = 29√2.
(√2+1)⁵ = 41 + 29√2
In-text Example 5: Binomial Theorem ka use karke prove karo ki 6n − 5n hamesha 25 se divide karne par remainder 1 deta hai, jab n ek positive integer ho.
6ⁿ likho (1+5)ⁿ ki tarah aur expand karo:
6ⁿ = Σr=0n ⁿCr 5r = 1 + 5n + Σr=2n ⁿCr 5r
r ≥ 2 wale saare terms mein 5²=25 ka factor common hai, isliye unka sum 25k likha ja sakta hai (k ek integer hai).
6ⁿ = 1 + 5n + 25k ⇒ 6ⁿ − 5n − 1 = 25k
Yani 6ⁿ − 5n ko 25 se divide karne par quotient badhta hai lekin remainder hamesha 1 rehta hai. Proved.
In-text Example 6: (x² − y)6 ke expansion ka general term likho.
a = x², b = −y, n = 6.
Tr+1 = &sup6;Cr (x²)6−r (−y)r = (−1)r &sup6;Cr x12−2r yr
In-text Example 7: (x/3 + 9y)10 ke expansion mein middle term nikalo.
n = 10 (even), isliye sirf ek middle term hota hai — (n/2 + 1)th term = 6th term, yaani r = 5.
T₆ = T₅ₓ₁ = ¹⁰C₅ (x/3)⁵ (9y)⁵
¹⁰C₅ = 252, (x/3)⁵ = x⁵/243, (9y)⁵ = 9⁵y⁵ = 59049y⁵. Aur 59049/243 = 243.
T₆ = 252 × 243 × x⁵y⁵ = 61236 x⁵y⁵
In-text Example 8: (x + y)9 ke expansion mein x6y3 ka coefficient nikalo.
General term: Tr+1 = &sup9;Cr x9−r yr. Chahiye 9−r=6 aur r=3, dono se r=3 milta hai — consistent.
Coefficient = &sup9;C₃ = 84
In-text Example 9: (x − 1/x)8 ke expansion mein x se independent term (constant term) nikalo.
Tr+1 = &sup8;Cr x8−r(−1/x)r = (−1)r &sup8;Cr x8−2r
x se independent matlab exponent zero: 8−2r=0 ⇒ r=4.
T₅ = (−1)⁴ &sup8;C₄ = 1 × 70 = 70
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q20)
Exercise 7.1, Q1: Expand (1 − 2x)5.
Coefficients ⁵Cr: 1,5,10,10,5,1. a=1, b=−2x.
(1−2x)⁵ = 1 − 5(2x) + 10(4x²) − 10(8x³) + 5(16x⁴) − 32x⁵
= 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵
Exercise 7.1, Q2: Expand (2/x − x/2)5.
a = 2/x, b = −x/2, coefficients 1,5,10,10,5,1.
T₁=32/x⁵, T₂=−40/x³, T₃=20/x, T₄=−5x, T₅=5x³/8, T₆=−x⁵/32
(2/x − x/2)⁵ = 32/x⁵ − 40/x³ + 20/x − 5x + (5/8)x³ − x⁵/32
Exercise 7.1, Q3: Expand (2x − 3)6.
a=2x, b=−3, coefficients 1,6,15,20,15,6,1.
= 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729
Exercise 7.1, Q4: Expand (x/3 + 1/x)5.
a=x/3, b=1/x, coefficients 1,5,10,10,5,1.
= x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵
Exercise 7.1, Q5: Expand (x + 1/x)6.
a=x, b=1/x, coefficients 1,6,15,20,15,6,1; power of x = 6−2r.
= x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶
Exercise 7.1, Q6: Using the Binomial Theorem, evaluate (96)3.
96 = 100 − 4.
(100−4)³ = 100³ − 3(100²)(4) + 3(100)(16) − 64
= 1,000,000 − 120,000 + 4,800 − 64 = 884,736
Exercise 7.1, Q7: Using the Binomial Theorem, evaluate (102)5.
102 = 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
= 11,040,808,032
Exercise 7.1, Q8: Using the Binomial Theorem, evaluate (101)4.
101 = 100 + 1.
(100+1)⁴ = 100⁴ + 4(100³) + 6(100²) + 4(100) + 1
= 100,000,000 + 4,000,000 + 60,000 + 400 + 1 = 104,060,401
Exercise 7.1, Q9: Using the Binomial Theorem, evaluate (99)5.
99 = 100 − 1.
(100−1)⁵ = 100⁵ − 5(100⁴) + 10(100³) − 10(100²) + 5(100) − 1
= 10,000,000,000 − 500,000,000 + 10,000,000 − 100,000 + 500 − 1
= 9,509,900,499
Exercise 7.1, Q10: Using Binomial Theorem, indicate which number is larger: (1.1)10000 ya 1000?
(1.1)10000 = (1+0.1)10000 ko expand karo:
(1+0.1)10000 = 1 + 10000(0.1) + (baaki sab positive terms)
Pehle do terms hi 1 + 1000 = 1001 de dete hain, aur agle saare terms bhi positive hain (kyunki (0.1)>0), isliye pura sum > 1001.
∴ (1.1)¹⁰⁰⁰⁰ > 1000
Exercise 7.1, Q11: Find (a+b)4 − (a−b)4. Hence evaluate (√3 + √2)4 − (√3 − √2)4.
Dono expansion likh ke odd-index terms (r=1,3) bachte hain, even-index terms cancel ho jaate hain:
(a+b)⁴ − (a−b)⁴ = 2[⁴C₁a³b + ⁴C₃ab³] = 2[4a³b + 4ab³] = 8ab(a²+b²)
a = √3, b = √2: ab = √6, a²+b² = 3+2 = 5.
(√3+√2)⁴ − (√3−√2)⁴ = 8√6(5) = 40√6
Exercise 7.1, Q12: Find (x+1)6 + (x−1)6. Hence or otherwise evaluate (√2+1)6 + (√2−1)6.
Is baar odd-index terms cancel hote hain, even-index terms double ho jaate hain:
(x+1)⁶+(x−1)⁶ = 2[x⁶+15x⁴+15x²+1]
x = √2: x²=2, x⁴=4, x⁶=8.
= 2[8 + 60 + 30 + 1] = 2(99) = 198
Exercise 7.1, Q13: Show that 9n+1 − 8n − 9 hamesha 64 se divisible hai, jab n koi positive integer ho.
9ⁿ⁺¹ = (1+8)ⁿ⁺¹ ko expand karo:
= Σr=0n+1 n+1Cr 8r = 1 + (n+1)(8) + Σr=2n+1 n+1Cr 8r
r ≥ 2 wale saare terms mein 8²=64 ka factor common hai, unka sum 64m likho:
9ⁿ⁺¹ = 1 + 8n + 8 + 64m = 9 + 8n + 64m
⇒ 9ⁿ⁺¹ − 8n − 9 = 64m, jo 64 se divisible hai. Proved.
Exercise 7.1, Q14: Prove that Σr=0n 3r ⁿCr = 4n.
Binomial theorem se (1+x)ⁿ = ΣⁿCrxr. x = 3 rakho:
(1+3)ⁿ = Σr=0n ⁿCr 3r
LHS = 4ⁿ, isliye Σr=0n 3r ⁿCr = 4ⁿ. Proved.
Miscellaneous Exercise, Q1: (x+1)n ke expansion mein (r−1)th, rth aur (r+1)th terms ke coefficients 1:3:5 ratio mein hain. n aur r nikalo.
(x+1)ⁿ mein kth term ka coefficient ⁿCk−1 hota hai. Isliye:
(r−1)th term coeff = ⁿCr−2, rth term coeff = ⁿCr−1, (r+1)th term coeff = ⁿCr
Diya hai ⁿCr−2 : ⁿCr−1 : ⁿCr = 1:3:5.
ⁿCr−1/ⁿCr−2 = 3 ⇒ (n−r+2)/(r−1) = 3 ⇒ n = 4r − 5 …(i)
ⁿCr/ⁿCr−1 = 5/3 ⇒ (n−r+1)/r = 5/3 ⇒ 3n − 8r + 3 = 0 …(ii)
(i) ko (ii) mein daalo: 3(4r−5) − 8r + 3 = 0 ⇒ 4r − 12 = 0 ⇒ r = 3, isliye n = 4(3)−5 = 7.
n = 7, r = 3
Miscellaneous Exercise, Q2: (3 + ax)9 ke expansion mein x² aur x³ ke coefficients equal hain. a nikalo.
Tr+1 = &sup9;Cr 39−r (ax)r.
x² ka coeff (r=2) = &sup9;C² 3⁵a² = 36 × 2187 × a² = 78732a²
x³ ka coeff (r=3) = &sup9;C³ 3⁶a³ = 84 × 729 × a³ = 61236a³
Dono equal: 78732a² = 61236a³ ⇒ a = 78732/61236 = 9/7 (a ≠ 0 maan ke divide kiya).
a = 9/7
Miscellaneous Exercise, Q3: Binomial theorem use karke (1+2x)6(1−x)7 ke product mein x⁵ ka coefficient nikalo.
(1+2x)⁶ = Σ⁶Ck2kxk, (1−x)⁷ = Σ⁷Cj(−1)jxj. x⁵ ka coefficient un pairs se aata hai jinme k+j=5:
k=0,j=5: 1×1 × 21×(−1) = −21
k=1,j=4: 6×2 × 35×1 = 420
k=2,j=3: 15×4 × 35×(−1) = −2100
k=3,j=2: 20×8 × 21×1 = 3360
k=4,j=1: 15×16 × 7×(−1) = −1680
k=5,j=0: 6×32 × 1×1 = 192
Sum: −21+420−2100+3360−1680+192 = 171.
x⁵ ka coefficient = 171
Miscellaneous Exercise, Q4: Agar a aur b distinct integers hain, prove karo ki (a−b), an−bn ka factor hai, jab n koi positive integer ho. [Hint: an = (a−b+b)n likh ke expand karo]
aⁿ = [(a−b) + b]ⁿ ko binomial theorem se expand karo:
aⁿ = (a−b)ⁿ + ⁿC₁(a−b)ⁿ⁻¹b + … + ⁿCn−1(a−b)bⁿ⁻¹ + bⁿ
bⁿ ko is-taraf le jao aur (a−b) common factor nikaalo:
aⁿ − bⁿ = (a−b)[(a−b)ⁿ⁻¹ + ⁿC₁(a−b)ⁿ⁻²b + … + ⁿCn−1bⁿ⁻¹]
Bracket ke andar sab integers hain, isliye poora expression ek integer × (a−b) hai — matlab (a−b), aⁿ−bⁿ ko exactly divide karta hai. Proved.
Miscellaneous Exercise, Q5: (√3 + √2)6 − (√3 − √2)6 evaluate karo.
Odd-index terms hi bachte hain (even cancel ho jaate hain), aur woh double ho jaate hain:
(a+b)⁶−(a−b)⁶ = 2[⁶C₁a⁵b + ⁶C₃a³b³ + ⁶C₅ab⁵] = 2[6a⁵b + 20a³b³ + 6ab⁵]
a=√3, b=√2: a⁵=9√3, b=√2, a³=3√3, b³=2√2, b⁵=4√2.
6a⁵b = 6(9√3)(√2) = 54√6; 20a³b³ = 20(3√3)(2√2) = 120√6; 6ab⁵ = 6(√3)(4√2) = 24√6
Sum inside = 198√6, so answer = 2 × 198√6 = 396√6
Miscellaneous Exercise, Q6: (a² + √(a²−1))4 + (a² − √(a²−1))4 ki value nikalo.
x = a², y = √(a²−1) maan lo. Odd-index terms cancel, even-index double ho jaate hain:
(x+y)⁴+(x−y)⁴ = 2[x⁴ + 6x²y² + y⁴]
x²=a⁴, y²=a²−1, x⁴=a⁸, y⁴=(a²−1)²=a⁴−2a²+1, 6x²y²=6a⁴(a²−1)=6a⁶−6a⁴.
Bracket = a⁸ + 6a⁶ − 6a⁴ + a⁴ − 2a² + 1 = a⁸ + 6a⁶ − 5a⁴ − 2a² + 1
Final answer = 2a⁸ + 12a⁶ − 10a⁴ − 4a² + 2
Important Equations — Ek Nazar Me
| Rule / Formula | Statement | Note |
|---|---|---|
| Binomial Theorem | (a+b)n = Σr=0n nCr an−rbr | n koi bhi positive integer |
| (a−b)n | Σr=0n (−1)r nCr an−rbr | signs alternate: +,−,+,−... |
| General term | Tr+1 = nCr an−rbr | (r+1)th term ke liye r use karo, r nahi |
| Number of terms | (a+b)n mein (n+1) terms | r = 0 se n tak |
| nCr formula | nCr = n! / (r!(n−r)!) | nCr = nCn−r |
| Pascal's Triangle relation | nCr + nCr−1 = n+1Cr | har row pichli row ke adjacent terms jodke banti hai |
| Middle term (n even) | ek hi middle term: (n/2 + 1)th term | r = n/2 |
| Middle terms (n odd) | do middle terms: ((n+1)/2)th aur ((n+3)/2)th | dono terms bindu ke aas-paas symmetric hote hain |
| Sum of coefficients | nC0+nC1+…+nCn = 2n | (1+1)n expand karke milta hai |
| Alternating sum | nC0−nC1+nC2−… = 0 | (1−1)n expand karke milta hai |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
Sign error in (a−b)n: students (a+b)n ka pattern seedha copy kar lete hain aur alternating +,−,+,−... signs bhool jaate hain, especially jab b khud negative ya fractional ho (jaise −x/2).
Term-index confusion: general term formula Tr+1 = nCran−rbr mein hai Tr+1, Tr nahi. "5th term" poochne par students r=5 le lete hain jabki sahi hai r=4.
Middle term rule mix-up: n even ho to sirf EK middle term hota hai ((n/2+1)th), aur n odd ho to DO middle terms hote hain — students yeh rule ulta yaad rakh lete hain.
nCr calculation errors: factorial expand karte waqt nCr=nCn−r shortcut use nahi karte, aur bade factorials manually multiply karte waqt galti kar dete hain.
Exponent laws mistakes: (2/x)r ya (−x/2)r jaise fractional/negative base ko simplify karte waqt power aur sign dono par error hoti hai — ek term ka sign sahi hota hai lekin coefficient galat, ya vice versa.
Coefficient vs term confusion: jab sawaal "coefficient of x6" poochta hai, students poora term (x6 sahit) likh dete hain jabki answer sirf numeric coefficient hona chahiye.
Board-Style Important Questions
- Practice Question 1: (a+b)12 ke expansion mein total kitne terms honge?
- Practice Question 2: (x − 1/x²)9 ke expansion mein x6 ka coefficient nikalo.
- Practice Question 3: Binomial theorem use karke prove karo ki 32n+2 − 8n − 9, sabhi n ∈ N ke liye 8 se divisible hai.
- Practice Question 4: (2x − x²/4)9 ke expansion mein middle term(s) nikalo.
- Practice Question 5: Agar (1+x)n ke expansion mein 5th, 6th aur 7th terms ke coefficients A.P. mein hain, to n nikalo.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths NCERT Solutions Chapter 7 Binomial Theorem mein kya-kya cover hota hai?
Teen sections: 7.1 Introduction, 7.2 Binomial Theorem for Positive Integral Indices (jisme Pascal's Triangle bhi aata hai), aur 7.3 General and Middle Terms. Exercise 7.1 mein expansion aur evaluate-type 14 questions hain, aur Miscellaneous Exercise mein 6 conceptual/proof-type questions hain — dono is spec mein step-by-step solve kiye gaye hain, jaise class 11 maths ncert solutions ke standard format mein hota hai.
Binomial Theorem ka main formula kya hai, aur formulas revision ke liye kaise download karein?
Core formula: Tr+1 = nCr an−rbr, jisse poora expansion, general term, aur middle term teeno nikalte hain — full list is spec ke "formulas" table mein hai. Quick revision ke liye class 11 maths formulas pdf download karke poore chapter ki cheat-sheet ek jagah rakh sakte ho.
Exercise 7.1 mein kitne questions hain aur unka difficulty pattern kaisa hai?
Exercise 7.1 mein 14 questions hain: Q1–5 pure expansion practice, Q6–10 "using binomial theorem evaluate" type (jaise 99&sup5;, 101⁴), aur Q11–14 proof/identity-type — thoda tough. Miscellaneous Exercise ke 6 questions inse bhi zyada conceptual hote hain (AP coefficients, divisibility, product expansion).
Middle term nikalne ka shortcut trick kya hai?
Sirf n dekho: n even ho to ek hi middle term hota hai — (n/2 + 1)th term. n odd ho to do middle terms hote hain — ((n+1)/2)th aur ((n+3)/2)th. Dono cases mein pehle r nikaalo, phir Tr+1 formula lagao.
Kya is chapter ke liye NCERT Exemplar ya 2026-27 syllabus mein koi extra topic aata hai?
Textbook ke andar scope wahi teen sections (7.1–7.3) tak hi hai — koi extra numbered topic nahi. Extra practice ke liye class 11 maths exemplar solutions dekh sakte ho (harder variety ke same-topic questions), aur agar confirm karna ho ki chapter list badla nahi hai to class 11 maths syllabus 2026-27 pdf ncert.nic.in se check kar sakte ho. Original text ke liye class 11 maths ncert book pdf download bhi kar sakte ho.
Binomial Theorem ke baad revision ke liye Class 11 Maths ke kaunse chapters priority pe rakhein?
Algebra thread continue karne ke liye class 11 maths chapter 8 sequences and series important questions dekho — dono chapters mein series/pattern-thinking common hai. Foundation revise karna ho to class 11 maths chapter 1 sets ncert solutions aur class 11 maths chapter 3 trigonometric functions ncert solutions se shuru karo. Coordinate geometry side ke liye class 11 maths straight lines ncert solutions aur class 11 maths conic sections notes, calculus intro ke liye class 11 maths limits and derivatives ncert solutions, aur class 11 maths probability important questions bhi exam-important hain. Overall strategy ke liye class 11 maths important questions with solutions aur (jab official numbers available ho) class 11 maths chapter wise weightage 2026-27 dekhna helpful rahega.
Class 11 Maths — Saare Chapters
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