source: ncert
topic: algebra
class_level: class_11
chapter: ch8
difficulty: intermediate
Class 11 | Ch8: Binomial Theorem BINOMIAL THEOREM: (a+b)^n = sum(r=0 to n) nCr × a^(n-r) × b^r. General term: T(r+1) = nCr × a^(n-r) × b^r. (r starts from 0.) PROPERTIES: Number of terms = n+1. Sum of binomial coefficients = 2^n (put a=b=1). Sum of odd-position coefficients = Sum of even-position = 2^(n-1). Binomial coefficients: nC0,nC1,...,nCn form Pascal's triangle. MIDDLE TERM: If n is even: middle term is T(n/2+1). One middle term. If n is odd: two middle terms T((n+1)/2) and T((n+3)/2). TERM INDEPENDENT OF x: find r such that power of x in T(r+1) = 0. GREATEST TERM: find r such that T(r+1)/T(r) >= 1. SOLVED EXAMPLES: Example 1: Expand (2x-3y)^4. T(r+1)=4Cr×(2x)^(4-r)×(-3y)^r. r=0: 4C0×16x^4=16x^4. r=1: 4C1×8x^3×(-3y)=-96x^3y. r=2: 4C2×4x^2×9y^2=216x^2y^2. r=3: 4C3×2x×(-27y^3)=-216xy^3. r=4: 4C4×81y^4=81y^4. Expansion: 16x^4-96x^3y+216x^2y^2-216xy^3+81y^4. Example 2: Find 5th term in (x+2)^8. T5=T(4+1): r=4. T5=8C4×x^4×2^4=70×x^4×16=1120x^4. Example 3: Term independent of x in (x+1/x)^10. T(r+1)=10Cr×x^(10-r)×(1/x)^r=10Cr×x^(10-2r). For independent of x: 10-2r=0 → r=5. T6=10C5=252. COMMON MISTAKES: General term T(r+1) has r starting from 0, NOT 1. For "5th term": r=4 (since T(r+1)=T5 means r+1=5, r=4). In (a-b)^n: alternate signs. T(r+1)=nCr×a^(n-r)×(-b)^r.