Math101learn.math101.caBinomial Theorem
The binomial theorem expands a nonnegative integer power of a sum: $(a+b)^n=\sum_{k=0}^n\binom nk a^{n-k}b^k$.
The theorem expands large powers efficiently and connects algebra with combinations, probability, calculus, and approximation.
Intuition and core definition
The binomial theorem expands a nonnegative integer power of a sum: $(a+b)^n=\sum_{k=0}^n\binom nk a^{n-k}b^k$. The coefficients count ways to choose which $k$ of the $n$ factors contribute $b$. Exponents on $a$ decrease while exponents on $b$ increase, and every term has total degree $n$.
Notation, language, and conditions
$\binom nk=\frac{n!}{k!(n-k)!}$ is read “$n$ choose $k$.” The expansion has $n+1$ terms before like-term coincidences. For $(a-b)^n$, treat the second term as $-b$, producing alternating signs according to $(-b)^k$.
Why this idea matters
The binomial theorem organizes every term of a power using combinatorial coefficients, avoiding repeated distribution while preserving exponent patterns.
A dependable method
- Identify the two binomial terms, including any coefficient or negative sign, and the integer exponent $n$.
- List coefficients from Pascal’s triangle or compute $\binom nk$.
- Write powers of the first term from $n$ down to $0$ and powers of the second from $0$ up to $n$.
- Apply coefficients and simplify signs and numerical powers.
- Check term count, total degree, endpoints, and a simple substitution.
Worked example
Representations and interpretation
Pascal’s triangle supplies coefficient rows, while a combinatorial selection diagram explains them: choosing $k$ copies of the second term among $n$ identical factors creates $\binom nk$ equal products.
Reasoning about variations
The theorem is not simply “raise each term”: $(a+b)^n\ne a^n+b^n$ for $n>1$ in general. For a requested single term, the general term $T_{k+1}=\binom nk a^{n-k}b^k$ avoids expanding everything.
Common mistakes
How to check your work
- Set variables to $1$ and compare the expansion with the original binomial value.
- Verify first and last terms are $a^n$ and $b^n$ including internal coefficients.
- Check each unsimplified term has exponent total $n$.
Practice
- Find the coefficient of $x^2$ in $(x+2)^4$.
- Expand $(a-b)^3$.
- How many terms are in the standard expansion of $(p+q)^7$?
Answers and brief solutions
Show answers
- $24$ The $x^2$ term is $\binom42x^2(2)^2=6\cdot4x^2$.
- $a^3-3a^2b+3ab^2-b^3$ Use coefficients $1,3,3,1$ and powers of $-b$.
- $8$ The index $k$ runs from $0$ through $7$.
Synthesis and transfer
To find one coefficient in a high power, select the corresponding combination and complementary exponents; checking the total degree catches an incorrectly indexed term.
In $(x+2)^6$, the $x^4$ term occurs when two of the six factors contribute $2$, so its coefficient is $\binom62 2^2=60$. The exponents of $x$ and $2$ always add to six, and consecutive binomial coefficients follow Pascal's triangle. A single-term calculation therefore needs neither the entire expansion nor repeated distribution. Replacing the plus by a minus alternates signs according to the selected power, while changing the constants affects powers but not the combinatorial counts. Summing the coefficients by substituting $x=1$ gives a useful whole-expansion check: the result must equal $3^6$.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Find the coefficient of $x^2$ in $(x+2)^4$.
- The $x^2$ term is $\binom42x^2(2)^2=6\cdot4x^2$.
End of lesson
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