Binomial Theorem Class 11 Notes

The expansion of $(a+b{)}^n$ has (n + 1) terms. The powers of a decrease from n to 0 while the powers of b increase from 0 to n. Each term's coefficient is the corresponding binomial coefficient ${}^n{C}_r$.

• For (1 + x)^n, set a = 1 to get a simpler form:
• $(1+x{)}^n={\sum }_{r=0}^n{}^n{C}_r{x}^r$
• .
• The general term in the expansion of (a + b)^n is the (r + 1)-th term.
• The sum of all binomial coefficients equals 2^n.

• Row 0: 1. Row 1: 1 1. Row 2: 1 2 1. Row 3: 1 3 3 1.
• Each row is symmetric: the first and last entries are 1, and interior entries are sums of the two above.
• Pascal's identity
• ${}^n{C}_r{+}^n{C}_{r-1}{=}^{n+1}{C}_r$
• is the rule that builds the triangle.

The (r + 1)-th term in the expansion of $(a+b{)}^n$ is obtained by substituting into the general term formula. For example, the 4th term (r = 3) in $(2x+3{)}^5$ is ${}^5{C}_3(2x{)}^2(3{)}^3$.

• When n is even (say n = 2m), there is a single middle term: the (m + 1)-th term, i.e. T_{m+1}.
• When n is odd (say n = 2m + 1), there are two middle terms: the (m + 1)-th and (m + 2)-th terms.

The middle term carries the largest binomial coefficient in the expansion. For (1 + x)^n with even n, the middle term is the numerically largest term when all binomial coefficients are positive.

• Symmetry:
• ${}^n{C}_r{=}^n{C}_{n-r}$
• — the r-th and (n − r)-th coefficients are equal.
• Sum of coefficients:
• ${\sum }_{r=0}^n{}^n{C}_r=2^n$
• (set a = b = 1).
• Alternating sum:
• ${\sum }_{r=0}^n(-1{)}^r{}^n{C}_r=0$
• (set a = 1, b = −1).
• Weighted sum:
• ${\sum }_{r=0}^{n}r{}^n{C}_r=n\cdot 2^{n-1}$
• (differentiate and set x = 1).

• Approximate (1 + x)^n ≈ 1 + nx for small x (first-order binomial approximation, useful in physics).
• Coefficient extraction: the coefficient of x^r in (1 + x)^n is simply
• ${}^n{C}_r$
• .
• For (a + bx)^n, the coefficient of x^r is
• ${}^n{C}_r{a}^{n-r}{b}^r$
• .
• Sums involving binomial coefficients can often be evaluated by choosing specific values of x in the expansion.

Example: Find the 5th term in the expansion of $(3x-2{)}^8$.

Solution: The 5th term has r = 4. $T_5={}^8{C}_4(3x{)}^4(-2{)}^4=70\times 81x^4\times 16=90720x^4$.

Example: Find the middle term in the expansion of $(x+\frac{1}{x}{)}^{10}$.

Solution: n = 10 (even), so the single middle term is T₆ with r = 5. $T_6={}^{10}{C}_5{x}^5{x}^{-5}=252$. The middle term is the constant 252.

Exercises (NCERT Q1–Q14)

• 1) Expand (1 − 2x)⁵.
• 2) Expand (2/x − x/2)⁵.
• 3) Expand (2x − 3)⁶.
• 4) Expand (x/3 + 1/x)⁵.
• 5) Expand (x + 1/x)⁶.
• 6) Using the binomial theorem, evaluate (96)³.
• 7) Using the binomial theorem, evaluate (102)⁵.
• 8) Using the binomial theorem, evaluate (101)⁴.
• 9) Using the binomial theorem, evaluate (99)⁵.
• 10) Using the binomial theorem, indicate which number is larger: (1.1)¹⁰⁰⁰⁰ or 1000.
• 11) Find (a + b)⁴ − (a − b)⁴. Hence evaluate (√3 + √2)⁴ − (√3 − √2)⁴.
• 12) Find (x + 1)⁶ + (x − 1)⁶. Hence evaluate (√2 + 1)⁶ + (√2 − 1)⁶.
• 13) Show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64, whenever n is a positive integer.
• 14) Prove that Σᵣ₌₀ⁿ 3ʳ ⁿCᵣ = 4ⁿ.

Answers at a glance

1. 1
   Q1 → 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.
2. 2
   Q2 → 32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.
3. 3
   Q3 → 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.
4. 4
   Q4 → x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.
5. 5
   Q5 → x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.
6. 6
   Q6 → 884736.
7. 7
   Q7 → 11040808032.
8. 8
   Q8 → 104060401.
9. 9
   Q9 → 9509900499.
10. 10
    Q10 → (1.1)¹⁰⁰⁰⁰.
11. 11
    Q11 → 8ab(a² + b²); hence 40√6.
12. 12
    Q12 → 2(x⁶ + 15x⁴ + 15x² + 1); hence 198.
13. 13
    Q13 → 9ⁿ⁺¹ − 8n − 9 = 64k (shown).
14. 14
    Q14 → Proved: LHS = 4ⁿ.

MCQ Questions

1 Mark Questions

1. 1
   Write the general term of (a + b)ⁿ. (Ans: T_(r+1) = nCr aⁿ⁻ʳ bʳ)
2. 2
   State the value of C(n, 0). (Ans: 1)
3. 3
   How many terms are there in (2x − 3y)⁹? (Ans: 10)
4. 4
   Write the coefficient of a²b in (a + b)³. (Ans: 3)

2 Mark Questions

1. 1
   Find the 4th term of (x + 2)⁵. (Ans: 5C3 x² 2³ = 80x²)
2. 2
   Find the term independent of x in (x² − 1/x)⁶. (Ans: term with exponent 0: 6C2 (−1)² = 15)
3. 3
   Find the middle term of (x + 1/x)⁸. (Ans: T₅ = 8C4 = 70)

3 Mark Questions

1. 1
   Using the binomial theorem, expand (2x − 1)⁴. (Ans: 16x⁴ − 32x³ + 24x² − 8x + 1)
2. 2
   Find the coefficient of x⁵ in the expansion of (1 + 2x)⁶. (Ans: 6C5 2⁵ = 192)
3. 3
   Evaluate (1.01)⁵ correct to two decimals using the binomial expansion. (Ans: 1 + 0.05 + 0.001 + 0.00001 ≈ 1.05)

4 Mark Questions

1. 1
   Prove that C(n, 0) + C(n, 1) + C(n, 2) + … + C(n, n) = 2ⁿ by putting x = 1 in (1 + x)ⁿ, and hence evaluate C(10, 0) + C(10, 1) + … + C(10, 10). (Ans: 2¹⁰ = 1024)
2. 2
   Find the term independent of x in (x² + 1/x)⁹, showing the general term and the condition on r. (Ans: r = 6 gives T₇ = 9C6 = 84)