This sheet covers the binomial theorem for positive integral indices — the expansion of \((a+b)^n\), the special cases \((x-y)^n\), \((1+x)^n\) and \((1-x)^n\), and the key properties of binomial coefficients. Each formula is grouped by topic, with the meaning of every symbol, guidance on when to use it, and original worked examples.
For the detailed derivation and complete explanations, refer to the Class 11 Maths notes page. The table below gives you every formula at a glance; scroll down for grouped lists, symbol meanings, and worked examples.
Formulas at a Glance
| Purpose (what you are finding) | Formula |
|---|---|
| Expansion of \((a+b)^n\) | \(\displaystyle (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\) |
| Expansion of \((x-y)^n\) | \(\displaystyle (x-y)^n = \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} x^{n-k} y^{k}\) |
| Expansion of \((1+x)^n\) | \(\displaystyle (1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k}\) |
| Expansion of \((1-x)^n\) | \(\displaystyle (1-x)^n = \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} x^{k}\) |
| Sum of all binomial coefficients | \(\displaystyle \sum_{k=0}^{n} \binom{n}{k} = 2^{n}\) |
| Alternating sum of binomial coefficients | \(\displaystyle \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0\) |
| Binomial coefficient (definition) | \(\displaystyle \binom{n}{r} = \frac{n!}{r!\,(n-r)!}, \; 0 \le r \le n\) |
All Formulas, Grouped by Topic
Binomial Theorem for Positive Integral Indices
For any positive integer \(n\),
\[ (a+b)^n = \binom{n}{0} a^{n} + \binom{n}{1} a^{n-1} b + \binom{n}{2} a^{n-2} b^{2} + \dots + \binom{n}{n-1} a b^{n-1} + \binom{n}{n} b^{n}. \]
This is equivalently written in sigma notation as
\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}. \]
The number of terms in the expansion is \((n+1)\). In each term, the sum of the exponents of \(a\) and \(b\) is \(n\).
The coefficients \(\binom{n}{r}\) are called binomial coefficients and are defined as
\[ \binom{n}{r} = \frac{n!}{r!\,(n-r)!}, \qquad 0 \le r \le n. \]
Special Cases
Case 1: Replace \(b\) by \(-y\) in the theorem:
\[ (x-y)^n = \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} x^{n-k} y^{k}. \]
Case 2: Put \(a = 1\) and \(b = x\):
\[ (1+x)^n = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^{2} + \dots + \binom{n}{n} x^{n}. \]
Case 3: Put \(a = 1\) and \(b = -x\):
\[ (1-x)^n = \binom{n}{0} – \binom{n}{1} x + \binom{n}{2} x^{2} – \dots + (-1)^{n} \binom{n}{n} x^{n}. \]
Properties of Binomial Coefficients
Substituting \(x = 1\) in \((1+x)^n\) gives the sum of all binomial coefficients:
\[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^{n}. \]
Substituting \(x = -1\) in \((1+x)^n\) gives the alternating sum:
\[ \binom{n}{0} – \binom{n}{1} + \binom{n}{2} – \dots + (-1)^{n} \binom{n}{n} = 0. \]
This is also written as \(\sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0\).
The coefficients in the expansions form Pascal’s triangle, where each entry is the sum of the two entries above it. The triangle is also represented using binomial coefficients: row \(n\) contains \(\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}\).
What Each Symbol Means
| Symbol | What it means | Unit / Nature |
|---|---|---|
| \(n\) | Positive integer index (the exponent of the binomial) | Integer (count) |
| \(r\) (or \(k\)) | Term number index, ranging from \(0\) to \(n\) | Integer (count) |
| \(a, b, x, y\) | Variables or numbers in the binomial | Real number (dimensionless) |
| \(\binom{n}{r}\) | Binomial coefficient — number of ways to choose \(r\) items from \(n\) | Positive integer |
| \(n!\) | Factorial of \(n\) (\(n \times (n-1) \times \dots \times 1\)) | Integer |
| \(\sum\limits_{k=0}^{n}\) | Summation from \(k=0\) to \(n\) | Operator |
When to Use Each Formula
| Formula | Use it when … |
|---|---|
| \((a+b)^n = \sum \binom{n}{k} a^{n-k} b^{k}\) | You need to expand a binomial to a positive integer power. The index \(n\) must be a positive integer. |
| \((x-y)^n = \sum (-1)^{k} \binom{n}{k} x^{n-k} y^{k}\) | The binomial is a difference (\(x-y\)). The signs alternate starting with \(+\). |
| \((1+x)^n = \sum \binom{n}{k} x^{k}\) | One term is 1, and you want a quick expansion for small \(x\) or to sum binomial coefficients. |
| \((1-x)^n = \sum (-1)^{k} \binom{n}{k} x^{k}\) | One term is 1 and the other is \(-x\). Signs alternate, all coefficients stay positive. |
| \(\sum \binom{n}{k} = 2^{n}\) | You need the total number of subsets of an \(n\)-element set, or to simplify an expression involving the sum of all binomial coefficients. |
| \(\sum (-1)^{k} \binom{n}{k} = 0\) | You need to prove identities or cancel terms in alternating sums. |
Worked Examples
Example 1: Expand \((2x + 3y)^5\)
- Step 1: Identify \(a = 2x\), \(b = 3y\), \(n = 5\).
- Step 2: Write the binomial theorem:
\[ (2x+3y)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} (3y)^{k} \]
Step 3: Compute each term:
\[ \begin{aligned} k=0: &\; \binom{5}{0} (2x)^{5} (3y)^{0} = 1 \cdot 32x^{5} = 32x^{5} \\ k=1: &\; \binom{5}{1} (2x)^{4} (3y)^{1} = 5 \cdot 16x^{4} \cdot 3y = 240x^{4}y \\ k=2: &\; \binom{5}{2} (2x)^{3} (3y)^{2} = 10 \cdot 8x^{3} \cdot 9y^{2} = 720x^{3}y^{2} \\ k=3: &\; \binom{5}{3} (2x)^{2} (3y)^{3} = 10 \cdot 4x^{2} \cdot 27y^{3} = 1080x^{2}y^{3} \\ k=4: &\; \binom{5}{4} (2x)^{1} (3y)^{4} = 5 \cdot 2x \cdot 81y^{4} = 810x y^{4} \\ k=5: &\; \binom{5}{5} (2x)^{0} (3y)^{5} = 1 \cdot 1 \cdot 243y^{5} = 243y^{5} \end{aligned} \]
Final answer: \( (2x+3y)^{5} = 32x^{5} + 240x^{4}y + 720x^{3}y^{2} + 1080x^{2}y^{3} + 810xy^{4} + 243y^{5} \).
Example 2: Evaluate \((1.02)^4\) using the binomial theorem (first three terms).
Step 1: Write \(1.02 = 1 + 0.02\).
So \((1.02)^4 = (1+0.02)^4\).
Step 2: Use \((1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k}\) with \(n=4\), \(x=0.02\).
\[ (1.02)^4 = \binom{4}{0}(1)^{4} + \binom{4}{1}(1)^{3}(0.02) + \binom{4}{2}(1)^{2}(0.02)^{2} + \cdots \]
Step 3: Compute the first three terms:
\[ \begin{aligned} \binom{4}{0} &= 1 \\ \binom{4}{1} \cdot 0.02 &= 4 \times 0.02 = 0.08 \\ \binom{4}{2} \cdot (0.02)^{2} &= 6 \times 0.0004 = 0.0024 \end{aligned} \]
Step 4: Add them: \(1 + 0.08 + 0.0024 = 1.0824\).
The remaining terms are much smaller (order \(10^{-5}\) and below).
Approximate answer: \((1.02)^4 \approx 1.0824\). (The exact value is \(1.08243216\).)
Example 3: Find the coefficient of \(x^{3}\) in the expansion of \((2x-1)^{7}\).
Step 1: Use the special case \((x-y)^n\) with \(x=2x\), \(y=1\), \(n=7\).
\[ (2x-1)^{7} = \sum_{k=0}^{7} \binom{7}{k} (2x)^{7-k} (-1)^{k} \]
- Step 1: The term containing \(x^{3}\) occurs when the exponent of \(x\) is 3, i.e., \(7-k = 3 \Rightarrow k = 4\).
- Step 2: Substitute \(k=4\):
\[ \binom{7}{4} (2x)^{3} (-1)^{4} = \binom{7}{4} \cdot 8x^{3} \cdot 1 \]
Step 4: Compute \(\binom{7}{4} = 35\).
So the term is \(35 \times 8x^{3} = 280x^{3}\).
Final answer: The coefficient of \(x^{3}\) is \(280\).
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Writing \((x-y)^n\) with all positive signs | The signs alternate: \((x-y)^n = \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} x^{n-k} y^{k}\). The first term is positive, second negative, and so on. | Test with \(n=2\): \((x-y)^2 = x^2 – 2xy + y^2\). Your expansion should match this pattern. |
| Forgetting the factor \(2^{n-k}\) or \(3^{k}\) when \(a\) or \(b\) is not 1 | Every term has \(a^{n-k} b^{k}\) — if \(a=2x\), then \(a^{n-k} = (2x)^{n-k} = 2^{n-k} x^{n-k}\). Always raise the entire coefficient and variable together. | In the expansion of \((2x)^3\), you should get \(8x^3\), not \(2x^3\). |
| Using the theorem for \(n\) that is not a positive integer | This chapter covers only positive integral indices. For negative or fractional indices, a different series (binomial series) is used in higher classes. | If the problem says \(n \in \mathbb{N}\), you are safe. Otherwise, don’t apply this formula. |
| Mistaking \(\binom{n}{r}\) for \(\frac{n}{r}\) instead of \(\frac{n!}{r!(n-r)!}\) | It is a combination number, not a fraction. For small \(n\) use Pascal’s triangle or the formula. | For \(\binom{5}{2}\), compute: \(\frac{5\times4}{2\times1} = 10\). Never write \(5/2\). |
| Omitting the constant term \(\binom{n}{0} a^n\) or \(\binom{n}{n} b^n\) | Both are always 1 and must be included. The expansion has \(n+1\) terms, from \(k=0\) to \(k=n\). | Count the terms: for \((a+b)^4\) you should get exactly 5 terms. |
Frequently Asked Questions
What is the binomial theorem for positive integral indices?
It gives a formula to expand \((a+b)^n\) when \(n\) is a positive integer. The expansion is \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\), and it contains \(n+1\) terms. The coefficients \(\binom{n}{k}\) are called binomial coefficients.
How do I find the general term in the expansion?
In \((a+b)^n\), the term corresponding to index \(k\) (starting from 0) is \(T_{k+1} = \binom{n}{k} a^{n-k} b^{k}\). For example, the third term (\(k=2\)) is \(T_3 = \binom{n}{2} a^{n-2} b^{2}\).
What is the sum of all binomial coefficients for a given \(n\)?
The sum of all binomial coefficients in the expansion of \((1+x)^n\) with \(x=1\) gives \(\sum_{k=0}^{n} \binom{n}{k} = 2^{n}\). This also equals the total number of subsets of an \(n\)-element set.
Why does the alternating sum of binomial coefficients equal zero?
Setting \(x=-1\) in \((1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^{k}\) gives \((1-1)^n = 0^n = 0 = \sum_{k=0}^{n} \binom{n}{k} (-1)^{k}\). This holds for any \(n \ge 1\).
Reference: NCERT Class 11 Mathematics textbook, chapter Binomial Theorem.
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More for this chapter:
- Binomial Theorem Notes
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Official source: download the NCERT textbook free from ncert.nic.in.