Binomial Theorem
Exercise 7.1 Solutions
$$ (a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r $$
$$ = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{n}b^n $$
[Image of Pascal’s Triangle]
Q1
1. Expand $(1 – 2x)^5$
Setup
$a=1, b=-2x, n=5$
Expand
$^5C_0(1)^5 + ^5C_1(1)^4(-2x) + ^5C_2(1)^3(-2x)^2 + ^5C_3(1)^2(-2x)^3 + ^5C_4(1)^1(-2x)^4 + ^5C_5(-2x)^5$
Values
$1(1) + 5(-2x) + 10(4x^2) + 10(-8x^3) + 5(16x^4) + 1(-32x^5)$
Simplify
$1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5$
Answer: $1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5$
Q2
2. Expand $(\frac{2}{x} – \frac{x}{2})^5$
Setup
$a = \frac{2}{x}, \quad b = -\frac{x}{2}, \quad n = 5$
Expand
$^5C_0(\frac{2}{x})^5 + ^5C_1(\frac{2}{x})^4(-\frac{x}{2}) + ^5C_2(\frac{2}{x})^3(-\frac{x}{2})^2 + ^5C_3(\frac{2}{x})^2(-\frac{x}{2})^3 + ^5C_4(\frac{2}{x})(-\frac{x}{2})^4 + ^5C_5(-\frac{x}{2})^5$
Calculate
$1(\frac{32}{x^5}) – 5(\frac{16}{x^4})(\frac{x}{2}) + 10(\frac{8}{x^3})(\frac{x^2}{4}) – 10(\frac{4}{x^2})(\frac{x^3}{8}) + 5(\frac{2}{x})(\frac{x^4}{16}) – 1(\frac{x^5}{32})$
Simplify
$\frac{32}{x^5} – \frac{40}{x^3} + \frac{20}{x} – 5x + \frac{5x^3}{8} – \frac{x^5}{32}$
Answer: $\frac{32}{x^5} – \frac{40}{x^3} + \frac{20}{x} – 5x + \frac{5x^3}{8} – \frac{x^5}{32}$
Q3
3. Expand $(2x – 3)^6$
Expand
$^6C_0(2x)^6 – ^6C_1(2x)^5(3) + ^6C_2(2x)^4(3)^2 – ^6C_3(2x)^3(3)^3 + ^6C_4(2x)^2(3)^4 – ^6C_5(2x)(3)^5 + ^6C_6(3)^6$
Coeffs
$1(64x^6) – 6(32x^5)(3) + 15(16x^4)(9) – 20(8x^3)(27) + 15(4x^2)(81) – 6(2x)(243) + 1(729)$
Result
$64x^6 – 576x^5 + 2160x^4 – 4320x^3 + 4860x^2 – 2916x + 729$
Answer: $64x^6 – 576x^5 + 2160x^4 – 4320x^3 + 4860x^2 – 2916x + 729$
Q4
4. Expand $(\frac{x}{3} + \frac{1}{x})^5$
Expand
$^5C_0(\frac{x}{3})^5 + ^5C_1(\frac{x}{3})^4(\frac{1}{x}) + ^5C_2(\frac{x}{3})^3(\frac{1}{x})^2 + ^5C_3(\frac{x}{3})^2(\frac{1}{x})^3 + ^5C_4(\frac{x}{3})(\frac{1}{x})^4 + ^5C_5(\frac{1}{x})^5$
Calculate
$\frac{x^5}{243} + 5(\frac{x^4}{81})(\frac{1}{x}) + 10(\frac{x^3}{27})(\frac{1}{x^2}) + 10(\frac{x^2}{9})(\frac{1}{x^3}) + 5(\frac{x}{3})(\frac{1}{x^4}) + \frac{1}{x^5}$
Simplify
$\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$
Answer: $\frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}$
Q5
5. Expand $(x + \frac{1}{x})^6$
Expand
$^6C_0 x^6 + ^6C_1 x^5 \frac{1}{x} + ^6C_2 x^4 \frac{1}{x^2} + ^6C_3 x^3 \frac{1}{x^3} + ^6C_4 x^2 \frac{1}{x^4} + ^6C_5 x \frac{1}{x^5} + ^6C_6 \frac{1}{x^6}$
Simplify
$x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$
Answer: $x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$
Evaluations
Using Binomial Theorem, evaluate the following:
6. $(96)^3$
$= (100 – 4)^3$
$= ^3C_0(100)^3 – ^3C_1(100)^2(4) + ^3C_2(100)(4)^2 – ^3C_3(4)^3$
$= 1000000 – 3(10000)(4) + 3(100)(16) – 64$
$= 1000000 – 120000 + 4800 – 64 = 884736$
$= ^3C_0(100)^3 – ^3C_1(100)^2(4) + ^3C_2(100)(4)^2 – ^3C_3(4)^3$
$= 1000000 – 3(10000)(4) + 3(100)(16) – 64$
$= 1000000 – 120000 + 4800 – 64 = 884736$
7. $(102)^5$
$= (100 + 2)^5$
$= 100^5 + 5(100^4)(2) + 10(100^3)(4) + 10(100^2)(8) + 5(100)(16) + 32$
$= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32$
$= 11040808032$
$= 100^5 + 5(100^4)(2) + 10(100^3)(4) + 10(100^2)(8) + 5(100)(16) + 32$
$= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32$
$= 11040808032$
8. $(101)^4$
$= (100 + 1)^4$
$= 100^4 + 4(100^3)(1) + 6(100^2)(1) + 4(100)(1) + 1$
$= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401$
$= 100^4 + 4(100^3)(1) + 6(100^2)(1) + 4(100)(1) + 1$
$= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401$
9. $(99)^5$
$= (100 – 1)^5$
$= 100^5 – 5(100^4) + 10(100^3) – 10(100^2) + 5(100) – 1$
$= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1$
$= 9509900499$
$= 100^5 – 5(100^4) + 10(100^3) – 10(100^2) + 5(100) – 1$
$= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1$
$= 9509900499$
Q10
10. Using Binomial Theorem, indicate which number is larger $(1.1)^{10000}$ or $1000$.
Step 1
Write $1.1$ as $(1 + 0.1)$.
Expand
$(1.1)^{10000} = (1 + 0.1)^{10000} = ^{10000}C_0 + ^{10000}C_1(0.1) + \text{other positive terms}$
Simplify
$= 1 + 10000 \times 0.1 + \dots$
$= 1 + 1000 + \dots$
$= 1 + 1000 + \dots$
Compare
$1001 + \text{positive terms} > 1000$
Answer: $(1.1)^{10000} > 1000$
Q11
11. Find $(a + b)^4 – (a – b)^4$. Hence, evaluate $(\sqrt{3} + \sqrt{2})^4 – (\sqrt{3} – \sqrt{2})^4$.
Expansion
$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$
$(a-b)^4 = a^4 – 4a^3b + 6a^2b^2 – 4ab^3 + b^4$
$(a-b)^4 = a^4 – 4a^3b + 6a^2b^2 – 4ab^3 + b^4$
Subtract
$(a+b)^4 – (a-b)^4 = 2(4a^3b + 4ab^3) = 8ab(a^2 + b^2)$
Substitute
Put $a = \sqrt{3}, b = \sqrt{2}$
Calc
$= 8(\sqrt{3})(\sqrt{2})[(\sqrt{3})^2 + (\sqrt{2})^2]$
$= 8\sqrt{6}(3 + 2) = 8\sqrt{6}(5)$
$= 8\sqrt{6}(3 + 2) = 8\sqrt{6}(5)$
Result: $40\sqrt{6}$
Q12
12. Find $(x + 1)^6 + (x – 1)^6$. Hence evaluate $(\sqrt{2} + 1)^6 + (\sqrt{2} – 1)^6$.
Logic
When adding $(A+B)^n + (A-B)^n$, odd terms cancel, even terms double.
Simplify
$= 2[x^6 + 15x^4 + 15x^2 + 1]$
Substitute
Put $x = \sqrt{2}$
Calc
$= 2[(\sqrt{2})^6 + 15(\sqrt{2})^4 + 15(\sqrt{2})^2 + 1]$
$= 2[8 + 15(4) + 15(2) + 1]$
$= 2[8 + 60 + 30 + 1] = 2(99)$
$= 2[8 + 15(4) + 15(2) + 1]$
$= 2[8 + 60 + 30 + 1] = 2(99)$
Result: 198
Q13
13. Show that $9^{n+1} – 8n – 9$ is divisible by 64.
Step 1
Write $9^{n+1}$ as $(1 + 8)^{n+1}$.
Expand
$= 1 + ^{n+1}C_1(8) + ^{n+1}C_2(8^2) + \dots + (8)^{n+1}$
Simplify
$= 1 + (n+1)8 + 64(^{n+1}C_2 + \dots)$
$= 1 + 8n + 8 + 64k$
$= 1 + 8n + 8 + 64k$
Subtract
Expression becomes: $(9 + 8n + 64k) – 8n – 9 = 64k$
Since result is $64k$, it is divisible by 64.
Q14
14. Prove that $\sum_{r=0}^{n} 3^r \binom{n}{r} = 4^n$.
Formula
Recall: $\sum_{r=0}^n \binom{n}{r} a^{n-r} b^r = (a+b)^n$
Compare
Here, term is $3^r \binom{n}{r}$. This implies $b=3$ and $a=1$ (since $1^{n-r}=1$).
Result
Therefore, sum $= (1 + 3)^n = 4^n$.
Hence Proved.