Binomial Theorem
Miscellaneous Exercise Solutions
Q1
1. If a and b are distinct integers, prove that $a – b$ is a factor of $a^n – b^n$.
[Hint: write $a^n = (a – b + b)^n$]
[Hint: write $a^n = (a – b + b)^n$]
Step 1
Let $a = (a – b) + b$. Then $a^n = [(a – b) + b]^n$.
Expand
$a^n = \binom{n}{0}(a-b)^n + \binom{n}{1}(a-b)^{n-1}b + \dots + \binom{n}{n-1}(a-b)b^{n-1} + \binom{n}{n}b^n$
Simplify
$a^n = (a-b)[\binom{n}{0}(a-b)^{n-1} + \dots + \binom{n}{n-1}b^{n-1}] + b^n$
Rearrange
$a^n – b^n = (a-b) \times (\text{Some Integer } k)$
Hence, $a-b$ is a factor of $a^n – b^n$.
Q2
2. Evaluate $(\sqrt{3} + \sqrt{2})^6 – (\sqrt{3} – \sqrt{2})^6$.
Formula
$(a+b)^6 – (a-b)^6 = 2[\binom{6}{1}a^5b + \binom{6}{3}a^3b^3 + \binom{6}{5}ab^5]$
Substitute
Put $a=\sqrt{3}$ and $b=\sqrt{2}$.
Term 1
$\binom{6}{1}(\sqrt{3})^5(\sqrt{2}) = 6(9\sqrt{3})(\sqrt{2}) = 54\sqrt{6}$
Term 2
$\binom{6}{3}(\sqrt{3})^3(\sqrt{2})^3 = 20(3\sqrt{3})(2\sqrt{2}) = 120\sqrt{6}$
Term 3
$\binom{6}{5}(\sqrt{3})(\sqrt{2})^5 = 6(\sqrt{3})(4\sqrt{2}) = 24\sqrt{6}$
Total
$2[54\sqrt{6} + 120\sqrt{6} + 24\sqrt{6}] = 2[198\sqrt{6}]$
Answer: $396\sqrt{6}$
Q3
3. Find the value of $(a^2 + \sqrt{a^2-1})^4 + (a^2 – \sqrt{a^2-1})^4$.
Formula
$(x+y)^4 + (x-y)^4 = 2[\binom{4}{0}x^4 + \binom{4}{2}x^2y^2 + \binom{4}{4}y^4]$
Substitute
Let $x = a^2$ and $y = \sqrt{a^2-1}$. Note $y^2 = a^2-1$.
Expand
$2 [ (a^2)^4 + 6(a^2)^2(a^2-1) + 1(a^2-1)^2 ]$
Simplify
$2 [ a^8 + 6a^4(a^2-1) + (a^4 – 2a^2 + 1) ]$
$= 2 [ a^8 + 6a^6 – 6a^4 + a^4 – 2a^2 + 1 ]$
$= 2 [ a^8 + 6a^6 – 6a^4 + a^4 – 2a^2 + 1 ]$
Answer: $2(a^8 + 6a^6 – 5a^4 – 2a^2 + 1)$
Q4
4. Find an approximation of $(0.99)^5$ using the first three terms of its expansion.
Rewrite
$(0.99)^5 = (1 – 0.01)^5$
Expand
$\binom{5}{0}(1)^5 – \binom{5}{1}(1)^4(0.01) + \binom{5}{2}(1)^3(0.01)^2 – \dots$
Terms
$1 – 5(0.01) + 10(0.0001)$ (First 3 terms only)
Calculate
$1 – 0.05 + 0.001 = 0.95 + 0.001$
Approximation: $0.951$
Q5
5. Expand using Binomial Theorem $(1 + \frac{x}{2} – \frac{2}{x})^4, x \ne 0$.
Group
Let $y = 1 + \frac{x}{2}$. Expression becomes $(y – \frac{2}{x})^4$.
Expand
$^4C_0 y^4 – ^4C_1 y^3(\frac{2}{x}) + ^4C_2 y^2(\frac{2}{x})^2 – ^4C_3 y(\frac{2}{x})^3 + ^4C_4(\frac{2}{x})^4$
Sub y
$(1+\frac{x}{2})^4 – \frac{8}{x}(1+\frac{x}{2})^3 + \frac{24}{x^2}(1+\frac{x}{2})^2 – \frac{32}{x^3}(1+\frac{x}{2}) + \frac{16}{x^4}$
Note
This involves expanding each sub-binomial term. A concise method groups terms as squares.
Alt Method
Note $(1 + \frac{x}{2} – \frac{2}{x})^4$. Just expanding terms yields:
$16/x^4 – 32/x^3 + 24/x^2 – 8/x + 1 + x + x^2/4 + \dots$
(Full expansion results in 9 terms)
(Full expansion results in 9 terms)
Q6
6. Find the expansion of $(3x^2 – 2ax + 3a^2)^3$.
Group
Let terms be $[3(x^2 + a^2) – 2ax]^3$. Let $P = 3(x^2+a^2)$ and $Q = 2ax$.
Expand
$(P – Q)^3 = P^3 – 3P^2Q + 3PQ^2 – Q^3$
Term P³
$27(x^2+a^2)^3 = 27(x^6 + a^6 + 3x^4a^2 + 3x^2a^4)$
Term 3P²Q
$3[9(x^4+a^4+2x^2a^2)](2ax) = 54ax(x^4+a^4+2x^2a^2)$
Combine
Simplify all expanded terms.
$27x^6 – 54ax^5 + 117a^2x^4 – 116a^3x^3 + 117a^4x^2 – 54a^5x + 27a^6$