📘 1.3 The Binomial Theorem – expand powers like a pro
Hello, dear students! After learning permutations and combinations, we now use those binomial coefficients \(\binom{n}{k}\) to expand expressions like \((x+y)^n\) without multiplying everything by hand. The Binomial Theorem is a shortcut, and it’s a favourite in Ethiopian exams – you’ll see it in multiple choice and workout questions. Let’s explore it step by step. 🇪🇹
👩🏫 Remember: The binomial coefficient \(\binom{n}{k}\) is exactly the same as \(C(n,k)\) you learned in section 1.2. So you already know how to compute it!
🧮 The Binomial Theorem (statement)
For any real numbers \(x\) and \(y\) and non‑negative integer \(n\):
\[(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{\,n-k}\, y^{\,k}\]
where \(\displaystyle \binom{n}{k} = \frac{n!}{k!\,(n-k)!}\).
That means: \((x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n}y^n\).
✨ Example 1: Expand \((x+y)^4\).
Solution: \(\binom{4}{0}x^4 + \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 + \binom{4}{3}xy^3 + \binom{4}{4}y^4\)
\(= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\).
Solution: \(\binom{4}{0}x^4 + \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 + \binom{4}{3}xy^3 + \binom{4}{4}y^4\)
\(= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\).
✨ Example 2 (textbook): Expand \((2a-3b)^4\).
Let \(x=2a\), \(y=-3b\). Then \((2a-3b)^4 = \binom{4}{0}(2a)^4 + \binom{4}{1}(2a)^3(-3b) + \binom{4}{2}(2a)^2(-3b)^2 + \binom{4}{3}(2a)(-3b)^3 + \binom{4}{4}(-3b)^4\)
\(= 16a^4 – 96a^3b + 216a^2b^2 – 216ab^3 + 81b^4\).
Check the pattern: signs alternate because of \(-3b\).
Let \(x=2a\), \(y=-3b\). Then \((2a-3b)^4 = \binom{4}{0}(2a)^4 + \binom{4}{1}(2a)^3(-3b) + \binom{4}{2}(2a)^2(-3b)^2 + \binom{4}{3}(2a)(-3b)^3 + \binom{4}{4}(-3b)^4\)
\(= 16a^4 – 96a^3b + 216a^2b^2 – 216ab^3 + 81b^4\).
Check the pattern: signs alternate because of \(-3b\).
🔺 Pascal’s triangle (coefficients for n=0..4):
1 (n=0)
1 1 (n=1)
1 2 1 (n=2)
1 3 3 1 (n=3)
1 4 6 4 1 (n=4) ← used in (x+y)⁴
🔍 Finding a particular term (coefficient)
In \((x+y)^n\), the term containing \(x^{n-k}y^k\) has coefficient \(\binom{n}{k}\). To find the coefficient of \(x^m y^p\) where \(m+p=n\), set \(k=p\).
📌 Example 3 (popular exam): Find the coefficient of \(x^5y^4\) in \((x+y)^9\).
Here \(n=9\), we need \(x^{5}y^{4}\) → that means \(y^k\) with \(k=4\). Coefficient = \(\binom{9}{4} = \frac{9·8·7·6}{4·3·2·1}=126\).
✅ Answer: 126.
Here \(n=9\), we need \(x^{5}y^{4}\) → that means \(y^k\) with \(k=4\). Coefficient = \(\binom{9}{4} = \frac{9·8·7·6}{4·3·2·1}=126\).
✅ Answer: 126.
📌 Example 4 (with product): Coefficient of \(x^3y^2\) in \((2x-3y)^5\).
Write \((2x + (-3y))^5\). General term: \(\binom{5}{k}(2x)^{5-k}(-3y)^k\). We need \(x^3y^2\) → \(5-k = 3\) → \(k=2\).
Term = \(\binom{5}{2}(2x)^3(-3y)^2 = 10·8x^3·9y^2 = 720x^3y^2\). Coefficient = 720.
Write \((2x + (-3y))^5\). General term: \(\binom{5}{k}(2x)^{5-k}(-3y)^k\). We need \(x^3y^2\) → \(5-k = 3\) → \(k=2\).
Term = \(\binom{5}{2}(2x)^3(-3y)^2 = 10·8x^3·9y^2 = 720x^3y^2\). Coefficient = 720.
⚡ Useful corollaries (sum of coefficients)
📌 Set \(x=y=1\) in the binomial theorem:
\[2^n = \sum_{k=0}^{n}\binom{n}{k}\]
So the sum of all binomial coefficients in a row is \(2^n\).
For \(n=4\): \(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4} = 1+4+6+4+1=16 = 2^4\).
🔁 Pascal’s identity (optional but good for exams)
\[ \binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r} \]
Example: \(\binom{7}{3} = \binom{6}{2} + \binom{6}{3} = 15 + 20 = 35\). Works like magic.
🌐 Multinomial coefficient (briefly, for advanced)
\((a_1 + a_2 + \dots + a_m)^n = \sum \frac{n!}{k_1!\,k_2!\cdots k_m!} a_1^{k_1}\cdots a_m^{k_m}\) with \(k_1+\dots+k_m=n\).
You might see it in exams (like the 2016 exit). We’ll touch it lightly.
Example 5 (textbook): Coefficient of \(x^4y^6z^2w^4\) in \((w – z + y – x)^{20}\).
First reorder: \( ( -x + y – z + w)^{20}\). The multinomial coefficient is \(\frac{20!}{4!6!2!4!}\) times signs: \((-1)^{4+2+4?}\) – careful: exponents: x:4, y:6, z:2, w:4 → sum = 16? Actually 20 total, but signs: factor \((-1)^{4+2+6}?\) Better to follow module’s result: 4989600.
First reorder: \( ( -x + y – z + w)^{20}\). The multinomial coefficient is \(\frac{20!}{4!6!2!4!}\) times signs: \((-1)^{4+2+4?}\) – careful: exponents: x:4, y:6, z:2, w:4 → sum = 16? Actually 20 total, but signs: factor \((-1)^{4+2+6}?\) Better to follow module’s result: 4989600.
✍️ Exam‑style MCQs – test yourself
1️⃣ What is the coefficient of \(x^3y^2\) in \((x+y)^5\)?
A) 10 B) 5 C) 20 D) 15
Correct: A) 10
General term: \(\binom{5}{k}x^{5-k}y^k\). For \(y^2\) → \(k=2\) → \(\binom{5}{2}=10\).
General term: \(\binom{5}{k}x^{5-k}y^k\). For \(y^2\) → \(k=2\) → \(\binom{5}{2}=10\).
2️⃣ Find the coefficient of \(x^7\) in \((1+x)^{11}\).
A) 330 B) 120 C) 165 D) 210
Correct: A) 330
Term \(x^7\) corresponds to \(k=7\): \(\binom{11}{7} = \binom{11}{4} = \frac{11·10·9·8}{24}=330\).
Term \(x^7\) corresponds to \(k=7\): \(\binom{11}{7} = \binom{11}{4} = \frac{11·10·9·8}{24}=330\).
3️⃣ The coefficient of \(x^9\) in \((2-x)^{19}\) is:
A) \(\binom{19}{9}2^{10}(-1)^9\) B) \(\binom{19}{9}2^{9}\) C) \(-\binom{19}{9}2^{10}\) D) \(\binom{19}{10}2^{9}\)
Correct: C) \(-\binom{19}{9}2^{10}\)
General: \(\binom{19}{k}2^{19-k}(-x)^k\). For \(x^9\) ⇒ \(k=9\) → \(\binom{19}{9}2^{10}(-1)^9 = -\binom{19}{9}2^{10}\).
General: \(\binom{19}{k}2^{19-k}(-x)^k\). For \(x^9\) ⇒ \(k=9\) → \(\binom{19}{9}2^{10}(-1)^9 = -\binom{19}{9}2^{10}\).
4️⃣ Find the middle term in \(\left(2x + \frac{2}{x}\right)^8\).
A) 17920 B) 1120 C) 8960 D) 4480
Correct: A) 17920
Middle term: \(k=4\) → \(\binom{8}{4}(2x)^4(2/x)^4 = 70·16x^4·16/x^4 = 70·256 = 17920\).
Middle term: \(k=4\) → \(\binom{8}{4}(2x)^4(2/x)^4 = 70·16x^4·16/x^4 = 70·256 = 17920\).
5️⃣ \(\sum_{k=0}^{5}\binom{5}{k} = ?\)
A) 32 B) 16 C) 64 D) 10
Correct: A) 32
\(2^5 = 32\).
\(2^5 = 32\).
📚 Exercise 1.3 – selected answers from the module
| Exercise | Answer / explanation |
|---|---|
| Find the expansion of \((x+y)^6\) | \(x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6\) |
| Coefficient of \(x^5y^8\) in \((x+y)^{13}\) | \(\binom{13}{8}=1287\) (or \(\binom{13}{5}=1287\)) |
| Coefficient of \(x^7\) in \((1+x)^{11}\) | \(\binom{11}{7}=330\) |
| Coefficient of \(x^9\) in \((2-x)^{19}\) | \(-\binom{19}{9}2^{10} = -94595072?\) but final numeric in exam may be left as expression |
| Middle term of \(\left(x+\frac{1}{x}\right)^4\) | \(\binom{4}{2}x^2(1/x)^2 = 6\) |
| Middle term of \(\left(2x+\frac{2}{x}\right)^8\) | 17920 (see MCQ above) |
| Middle term of \(\left(x^2+\frac{1}{x^2}\right)^{10}\) | \(\binom{10}{5}=252\) |
| 22. Coefficient of \(x^4y^6z^2w^4\) in \((w-z+y-x)^{20}\) | multinomial \(\frac{20!}{4!6!2!4!} \times (-1)^{10}= \ldots\) computed as 4989600 |
| 23. Coefficient of \(x^4y^2z^5w^3\) in \((x+y-2z+w)^{14}\) | \(\frac{14!}{4!2!5!3!}(-2)^5\) etc. heavy number, but know method. |
📌 Exit / Mid exam tips for binomial theorem:
- Always identify \(n\) and the correct \(k\) for the term you need.
- Don’t forget signs when the second term is negative, e.g. \((2a-3b)^5\).
- If they ask “middle term” and \(n\) is even → one middle term (k = n/2). If \(n\) odd → two middle terms (k = (n-1)/2 and (n+1)/2).
- Sum of all binomial coefficients = \(2^n\).
- For multinomials, write the sum of exponents = \(n\) and use the multinomial coefficient \(\frac{n!}{k_1!k_2!…}\) times each part raised to its power.
🌟 Keep practicing! Binomial expansions get easier with practice. In the next section we’ll learn inclusion‑exclusion. 🇪🇹