Binomial Theorem — Practice

Question 1

Write down row 6 of Pascal's triangle, and state the value of ⁶C₄.

Accepted Answer

Row 6 is 1, 6, 15, 20, 15, 6, 1 (entry r is ⁶Cᵣ). The entry at r=4 is 15, so ⁶C₄=15. Answer: Row 6 is 1,6,15,20,15,6,1 and ⁶C₄=15.

Question 2

Expand (x - 1)^4 using the Binomial Theorem.

Accepted Answer

Row 4 of Pascal's triangle is 1,4,6,4,1, with alternating signs since b=-1: x^4-4x^3+6x^2-4x+1. Answer: (x-1)^4 = x^4-4x^3+6x^2-4x+1.

Question 3

Expand (2x + 3)^5 using the Binomial Theorem.

Accepted Answer

Row 5 is 1,5,10,10,5,1. (2x)^5 + 5(2x)^4(3) + 10(2x)^3(3)^2 + 10(2x)^2(3)^3 + 5(2x)(3)^4 + (3)^5 = 32x^5+240x^4+720x^3+1080x^2+810x+243. Answer: (2x+3)^5 = 32x^5+240x^4+720x^3+1080x^2+810x+243.

Question 4

Using the Binomial Theorem, evaluate (102)^4.

Accepted Answer

Write 102=100+2. (100+2)^4 = ⁴C₀(100)^4+⁴C₁(100)^3(2)+⁴C₂(100)^2(2)^2+⁴C₃(100)(2)^3+⁴C₄(2)^4 = 100000000+8000000+240000+3200+16 = 108243216. Answer: (102)^4 = 108243216.

Question 5

If the sum of the binomial coefficients in the expansion of (1+x)^7 is required, find its value.

Accepted Answer

Setting x=1 in (1+x)^n gives 2^n as the sum of all coefficients. For n=7: 2^7=128. Answer: The sum of the coefficients is 128.

Question 6

Find the sum of the coefficients at the even positions (⁸C₀+⁸C₂+⁸C₄+⁸C₆+⁸C₈) in the expansion of (1+x)^8.

Accepted Answer

Adding the x=1 values of (1+x)^n and (1-x)^n cancels the odd-position terms, leaving the even-position sum equal to 2^(n-1). For n=8: 2^7=128. Answer: The even-position coefficients sum to 128.

Question 7

Using the first three terms of its binomial expansion, find an approximate value of (1.01)^6.

Accepted Answer

(1+0.01)^6 ≈ ⁶C₀ + ⁶C₁(0.01) + ⁶C₂(0.01)^2 = 1 + 6(0.01) + 15(0.0001) = 1+0.06+0.0015 = 1.0615. Answer: (1.01)^6 is approximately 1.0615.

Question 8

Using the Binomial Theorem, show that 7^n - 6n - 1 is divisible by 36 for every positive integer n.

Accepted Answer

Write 7^n=(1+6)^n = nC0+nC1(6)+nC2(6)^2+...+nCn(6)^n = 1+6n+36(nC2+6.nC3+...+6^(n-2).nCn). So 7^n-6n-1 = 36(nC2+6.nC3+...+6^(n-2).nCn), a multiple of 36. Answer: 7^n-6n-1 is always divisible by 36.

Question 9

Find (x+1)^5 - (x-1)^5. Hence evaluate (√3+1)^5 - (√3-1)^5.

Accepted Answer

(x+1)^5=x^5+5x^4+10x^3+10x^2+5x+1 and (x-1)^5=x^5-5x^4+10x^3-10x^2+5x-1. Subtracting cancels the even-r terms, leaving (x+1)^5-(x-1)^5=10x^4+20x^2+2. At x=√3: 10(9)+20(3)+2=90+60+2=152. Answer: (√3+1)^5-(√3-1)^5 = 152.

Practice — Binomial Theorem

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities