Chapter 7 - Binomial Theorem

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Topper

1. Introduction - Expanding (a+b)^n

Binomial theorem gives formula to expand (a+b)^n where n is positive integer. Instead of multiplying (a+b)×(a+b)×... n times, we use nCr coefficients.

Example (a+b)^2 = a^2+2ab+b^2, (a+b)^3 = a^3+3a^2b+3ab^2+b^3. Coefficients 1,2,1 and 1,3,3,1 come from Pascal triangle.

General expansion: (a+b)^n = nC0 a^n b^0 + nC1 a^{n-1} b^1 + nC2 a^{n-2} b^2 + ... + nCn a^0 b^n. Total n+1 terms. Power of a decreases n to 0, power of b increases 0 to n.

Example (x+2)^3 = 3C0 x^3 2^0 + 3C1 x^2 2^1 + 3C2 x^1 2^2 + 3C3 x^0 2^3 = 1×x^3×1 + 3×x^2×2 + 3×x×4 +1×8 = x^3+6x^2+12x+8.

Binomial theorem expand (a+b)^n n positive integer instead of multiplying n times use nCr | Example (a+b)^2=a^2+2ab+b^2 (a+b)^3=a^3+3a^2b+3ab^2+b^3 Coefficients 1,2,1 and 1,3,3,1 Pascal | General (a+b)^n = nC0 a^n b^0 + nC1 a^{n-1} b^1 + ... + nCn a^0 b^n Total n+1 terms Power a decreases n to 0 power b increases 0 to n | Example (x+2)^3 = 3C0 x^3 2^0 + 3C1 x^2 2^1 + 3C2 x^1 2^2 + 3C3 x^0 2^3 = x^3+6x^2+12x+8

Introduction - Expanding (a+b)^n

2. Pascal Triangle - Pattern of Coefficients

Pascal triangle is triangular array where each number is sum of two numbers above it. Top Row 0: 1, Row 1: 1 1, Row 2: 1 2 1, Row 3: 1 3 3 1, Row 4: 1 4 6 4 1, Row 5: 1 5 10 10 5 1.

Row n gives coefficients of (a+b)^n expansion. Example Row 2: 1 2 1 coefficients of (a+b)^2 =1·a^2+2·ab+1·b^2. Row 3: 1 3 3 1 coefficients of (a+b)^3.

Each entry is nCr. Row 4: 4C0=1 4C1=4 4C2=6 4C3=4 4C4=1 → 1 4 6 4 1.

Properties: Edges always 1, symmetry each row reads same forward backward, sum of Row n = 2^n. Example Row 4 sum 1+4+6+4+1=16=2^4.

Practical use: Quickly find coefficients without calculating nCr each time just build triangle.

Pascal triangle triangular array each number sum two above Top Row 0:1 Row1:1 1 Row2:1 2 1 Row3:1 3 3 1 Row4:1 4 6 4 1 Row5:1 5 10 10 5 1 | Row n coefficients (a+b)^n Row2 1 2 1 coefficients (a+b)^2 =1·a^2+2·ab+1·b^2 Row3 1 3 3 1 coefficients (a+b)^3 | Each entry nCr Row4 4C0=1 4C1=4 4C2=6 4C3=4 4C4=1 →1 4 6 4 1 | Properties Edges always 1 symmetry reads same forward backward sum Row n =2^n Example Row4 sum 1+4+6+4+1=16=2^4 | Practical Quickly find coefficients without calculating nCr build triangle

Pascal Triangle - Coefficients Pattern

3. Binomial Expansion Formula (a+b)^n = Σ nCr a^{n-r} b^r

Formal statement: (a+b)^n = Σ_{r=0}^{n} nCr a^{n-r} b^r where Σ sum r from 0 to n.

Expanded: nC0 a^n + nC1 a^{n-1} b + nC2 a^{n-2} b^2 + ... + nC(n-1) a^1 b^{n-1} + nCn b^n.

Number of terms n+1. Example n=4 → 5 terms.

Binomial coefficient nCr = n!/(r!(n-r)!) read n choose r.

Example (x+y)^4 = 4C0 x^4 +4C1 x^3 y +4C2 x^2 y^2 +4C3 x y^3 +4C4 y^4 =1·x^4+4·x^3 y+6·x^2 y^2+4·x y^3+1·y^4.

Example (2a-3b)^3 = (2a+(-3b))^3 = 3C0 (2a)^3 +3C1 (2a)^2 (-3b) +3C2 (2a)(-3b)^2 +3C3 (-3b)^3 =8a^3+3·4a^2·(-3b)+3·2a·9b^2+(-27b^3)=8a^3-36a^2b+54ab^2-27b^3.

Trick: For (a-b)^n replace b by -b alternate signs + - + -.

Formal (a+b)^n = Σ_{r=0}^{n} nCr a^{n-r} b^r Σ sum r 0 to n | Expanded nC0 a^n + nC1 a^{n-1} b + ... + nCn b^n Number terms n+1 Example n=4 →5 terms | Binomial coefficient nCr=n!/(r!(n-r)!) read n choose r | Example (x+y)^4 =4C0 x^4+4C1 x^3 y+4C2 x^2 y^2+4C3 x y^3+4C4 y^4= x^4+4x^3 y+6x^2 y^2+4x y^3+y^4 | Example (2a-3b)^3 =(2a+(-3b))^3 =3C0 (2a)^3+3C1 (2a)^2(-3b)+3C2(2a)(-3b)^2+3C3(-3b)^3=8a^3-36a^2b+54ab^2-27b^3 | Trick For (a-b)^n replace b by -b alternate signs + - + -

Expansion Formula (a+b)^n

4. General Term Tr+1 = nCr a^{n-r} b^r and Middle Term

General term is (r+1)th term Tr+1 = nCr a^{n-r} b^r where r from 0 to n. So first term T1 when r=0: nC0 a^n, second term T2 when r=1: nC1 a^{n-1} b, etc.

Use to find specific term without expanding full.

Example Find 3rd term of (x+2)^5: r+1=3 → r=2 Tr+1 =5C2 x^{5-2} 2^2 =10 × x^3 ×4=40x^3. So third term is 40x^3.

Middle term: If n even, n=2m even, only one middle term is (n/2+1)th term = (m+1)th term. Example n=4 even middle term is (4/2+1)=3rd term.

If n odd, n=2m+1 odd, two middle terms are (m+1)th and (m+2)th terms = (n+1)/2 and (n+3)/2 th terms. Example n=5 odd middle terms are 3rd and 4th terms (5+1)/2=3 and (5+3)/2=4.

Example (x+1)^4 n=4 even middle term 3rd term =4C2 x^2 1^2 =6x^2. (x+1)^5 n=5 odd middle terms 3rd and 4th: 5C2 x^3=10x^3 and 5C3 x^2=10x^2.

General term (r+1)th term Tr+1=nCr a^{n-r} b^r r 0 to n So first T1 r=0 nC0 a^n second T2 r=1 nC1 a^{n-1} b etc | Use to find specific term without expanding full | Example Find 3rd term (x+2)^5 r+1=3 r=2 Tr+1=5C2 x^{3} 2^2=10×x^3×4=40x^3 third term 40x^3 | Middle term If n even n=2m even only one middle term (n/2+1)th = (m+1)th Example n=4 even middle (4/2+1)=3rd | If n odd n=2m+1 odd two middle terms (m+1)th and (m+2)th = (n+1)/2 and (n+3)/2 th Example n=5 odd middle 3rd and 4th (5+1)/2=3 and (5+3)/2=4 | Example (x+1)^4 n=4 even middle 3rd =4C2 x^2=6x^2 (x+1)^5 n=5 odd middle 3rd and 4th 5C2 x^3=10x^3 and 5C3 x^2=10x^2

General Term Tr+1 and Middle Term

5. Properties - nCr Identities Sum of Coefficients

Important identities:

Symmetry: nCr = nC(n-r) choosing r same as choosing n-r to leave. Example 5C2=5C3=10.

Sum of coefficients: nC0+nC1+nC2+...+nCn =2^n. Put a=1 b=1 in (1+1)^n=2^n. Example n=3 sum 1+3+3+1=8=2^3.

Alternating sum: nC0-nC1+nC2-...+(-1)^n nCn=0. Put a=1 b=-1 (1-1)^n=0.

Sum even indices = sum odd indices =2^{n-1}. Example n=4 even sum nC0+nC2+nC4=1+6+1=8=2^3 odd sum nC1+nC3=4+4=8=2^3.

Pascal identity: nCr = (n-1)C(r-1)+(n-1)Cr each entry sum two above in Pascal triangle.

nC0=1 nCn=1 nC1=n.

Symmetry nCr=nC(n-r) choosing r same choosing n-r leave Example 5C2=5C3=10 | Sum coefficients nC0+nC1+...+nCn=2^n Put a=1 b=1 (1+1)^n=2^n Example n=3 sum 1+3+3+1=8=2^3 | Alternating sum nC0-nC1+nC2-...+(-1)^n nCn=0 Put a=1 b=-1 (1-1)^n=0 | Sum even = sum odd =2^{n-1} Example n=4 even sum nC0+nC2+nC4=1+6+1=8=2^3 odd sum nC1+nC3=4+4=8=2^3 | Pascal identity nCr=(n-1)C(r-1)+(n-1)Cr each entry sum two above Pascal | nC0=1 nCn=1 nC1=n

Properties - nCr Identities

6. Practical Applications - Approximation Probability

Approximation: For small x, (1+x)^n ≈1+nx when x small and n large? Actually binomial approximation for small x: (1+x)^n ≈1+nx+ n(n-1)/2 x^2 + ... useful to approximate. Example (1.01)^5 ≈1+5×0.01=1.05 approximate actual 1.051... close.

Example (0.99)^5 = (1-0.01)^5 ≈1-5×0.01=0.95 approximate actual 0.95099... close.

Probability: Tossing n coins number of ways to get r heads = nCr /2^n. Example 3 coins toss probability 2 heads =3C2 /8 =3/8.

Engineering: Signal processing, error correcting codes uses binomial coefficients.

Combination counting: Number of subsets of n items =2^n same as sum nCr.

Example: 10% price increase twice? (1+0.1)^2 =1+2×0.1+0.1^2=1.21 →21% total increase not 20% because of binomial second term.

Interest: Compound interest (1+r)^n expansion gives interest each year.

Approximation For small x (1+x)^n ≈1+nx when x small n large binomial approximation small x (1+x)^n≈1+nx+ n(n-1)/2 x^2 useful approximate Example (1.01)^5≈1+5×0.01=1.05 actual 1.051 close | Example (0.99)^5=(1-0.01)^5≈1-5×0.01=0.95 actual 0.95099 close | Probability Tossing n coins ways get r heads =nCr/2^n Example 3 coins probability 2 heads =3C2/8=3/8 | Engineering Signal processing error correcting codes uses binomial coefficients | Combination counting Number subsets n items =2^n sum nCr | Example 10% price increase twice (1+0.1)^2=1+2×0.1+0.1^2=1.21 →21% total increase not 20% because binomial second term | Interest Compound interest (1+r)^n expansion gives interest each year

Practical Applications

7. Worked Examples - Step by Step

Example 1: Expand (x+2)^4. n=4. (x+2)^4 =4C0 x^4 2^0 +4C1 x^3 2^1 +4C2 x^2 2^2 +4C3 x^1 2^3 +4C4 x^0 2^4 =1·x^4·1 +4·x^3·2 +6·x^2·4 +4·x·8 +1·16 =x^4+8x^3+24x^2+32x+16.

Example 2: Find coefficient of x^2 in (x+3)^5. General term Tr+1=5Cr x^{5-r} 3^r. Need power of x 2 →5-r=2 →r=3. So coefficient =5C3 ×3^3 =10×27=270. Term 270x^2.

Example 3: Find middle term of (x+1)^6 n=6 even middle term 4th term (6/2+1=4). T4 r=3 6C3 x^{3} 1^3 =20x^3.

Example 4: Find term independent of x in (x+1/x)^6? General term 6Cr x^{6-r} (1/x)^r =6Cr x^{6-r} x^{-r} =6Cr x^{6-2r}. Independent means power 0 →6-2r=0 →r=3. Term =6C3=20 independent term 20.

Example 1 Expand (x+2)^4 n=4 (x+2)^4=4C0 x^4 2^0+4C1 x^3 2^1+4C2 x^2 2^2+4C3 x^1 2^3+4C4 x^0 2^4=1·x^4·1+4·x^3·2+6·x^2·4+4·x·8+1·16=x^4+8x^3+24x^2+32x+16 | Example 2 Find coefficient x^2 in (x+3)^5 General Tr+1=5Cr x^{5-r} 3^r Need power x 2 →5-r=2 r=3 Coefficient =5C3×3^3=10×27=270 Term 270x^2 | Example 3 Find middle term (x+1)^6 n=6 even middle 4th term (6/2+1=4) T4 r=3 6C3 x^3 1^3=20x^3 | Example 4 Find term independent x in (x+1/x)^6 General 6Cr x^{6-r} (1/x)^r=6Cr x^{6-r} x^{-r}=6Cr x^{6-2r} Independent power 0 →6-2r=0 r=3 Term=6C3=20 independent 20

Worked Examples Step by Step

8. Formulas Short Tricks - Quick Revision

Must remember:

(a+b)^n = nC0 a^n + nC1 a^{n-1} b + ... + nCn b^n n+1 terms.

nCr = n!/(r!(n-r)!) = nC(n-r).

nC0=1 nCn=1 nC1=n.

General term Tr+1 = nCr a^{n-r} b^r.

Middle term: n even one middle (n/2+1)th term, n odd two middle (n+1)/2 and (n+3)/2 th terms.

Sum nC0+nC1+...+nCn=2^n.

Alternating sum nC0-nC1+...=0.

Sum even = sum odd =2^{n-1}.

Pascal nCr=(n-1)C(r-1)+(n-1)Cr.

(a-b)^n = nC0 a^n - nC1 a^{n-1} b + nC2 a^{n-2} b^2 - ... alternate signs.

For small x (1+x)^n ≈1+nx approximation.

Quick reference: Row 0 1, Row1 1 1, Row2 1 2 1, Row3 1 3 3 1, Row4 1 4 6 4 1, Row5 1 5 10 10 5 1.

Must remember (a+b)^n = nC0 a^n + nC1 a^{n-1} b + ... + nCn b^n n+1 terms | nCr=n!/(r!(n-r)!)=nC(n-r) | nC0=1 nCn=1 nC1=n | General Tr+1=nCr a^{n-r} b^r | Middle n even one middle (n/2+1)th n odd two middle (n+1)/2 and (n+3)/2 th | Sum nC0+...+nCn=2^n | Alternating sum 0 | Sum even = sum odd =2^{n-1} | Pascal nCr=(n-1)C(r-1)+(n-1)Cr | (a-b)^n = nC0 a^n - nC1 a^{n-1} b + ... alternate signs | For small x (1+x)^n≈1+nx approximation | Quick reference Row0 1 Row1 1 1 Row2 1 2 1 Row3 1 3 3 1 Row4 1 4 6 4 1 Row5 1 5 10 10 5 1

Formulas Short Tricks