Binomial Theorem — Chapter Test | Class 11 IMO

Q1

The coefficient of x³ in (1 + 3x)⁴ is:

Coefficient = ⁴C₃×3³=4×27=108.

Q2

Which row of Pascal's Triangle contains the coefficients of (a + b)⁶?

Coefficients are 1, 6, 15, 20, 15, 6, 1 which appear in the 7th row if counting first row as row 1.

Q3

The coefficient of x³ in (x − 2)⁵ is:

Coefficient=⁵C₂×(−2)²=10×4=40.

Q4

The coefficient of x⁴ in (1 + x)⁸ + (1 + x)⁷ is:

⁸C₄+⁷C₄=70+35=105.

Q5

The middle terms in the expansion of (x + y)¹¹ are:

For an odd power n = 11, there are 12 terms. The two middle terms are the (11+1)/2 = 6th and (11+3)/2 = 7th terms.

Q6

The term independent of x in the expansion of (x + 1/x)¹⁰ is:

T_(r+1) = 10Cr x^(10−r) (x⁻¹)ʳ = 10Cr x^(10−2r). For x⁰, 10−2r = 0 → r = 5. Term = 10C5 = 252.

Q7

According to Pascal's rule, 10C4 + 10C5 is equal to:

Pascal's rule states nCr + nC(r−1) = (n+1)Cr. Therefore, 10C5 + 10C4 = 11C5.

Q8

Using binomial approximation, the value of (0.99)⁴ is approximately:

(1 − 0.01)⁴ ≈ 1 − 4(0.01) = 1 − 0.04 = 0.96.

Q9

A population grows by 1% annually. Using approximation, (1.01)¹⁰ is about:

(1+x)¹⁰≈1+10x=1.10.

Q10

If the middle term of (x + 1/x)ⁿ is 924, find the value of n.

The middle term is nC(n/2). We know 12C6 = 924. Thus, n = 12.

Q11

Find the 5th term in the expansion of (x − 2y)⁷.

T_5 = T_(4+1) = 7C4 (x)³ (−2y)⁴. 7C4 = 35. (−2y)⁴ = 16y⁴. 35 × 16 = 560. Term is 560 x³ y⁴.

Q12

If nC(n−1) = 15, then n is:

nC(n−1) is the same as nC1, which equals n. Thus, n = 15.

Q13

Find the sum of the coefficients in the expansion of (x − 2y + 3z)⁵.

Put x = 1, y = 1, z = 1. The expression becomes (1 − 2 + 3)⁵ = 2⁵ = 32.

Q14

The expansion of (1 + x)³ begins with:

Using binomial theorem: 1 + 3x + 3x² + x³.

Q15

Using binomial theorem, an investor estimates the future value of a stock growing at 0.5% over 200 days as (1.005)²⁰⁰. The approximate first-order multiplier is:

(1 + 0.005)²⁰⁰ ≈ 1 + 200(0.005) = 1 + 1 = 2.00.