Binomial Theorem is the habit of expanding $(x+y)^n$ and reading off the coefficient or term you need. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
The expansion of $(x+y)^n$ is a sum of terms whose coefficients are the binomial coefficients, so you can write down any specific term without multiplying the binomial out.
Why It Works
1
$(x + y)^n$ is the product of $n$ identical factors $(x + y)$; expanding it, each term picks either $x$ or $y$ from every factor.
2
A term with $y$ chosen from exactly $k$ of the factors (and $x$ from the other $n - k$) equals $x^{\,n-k}y^{\,k}$.
3
The number of ways to choose which $k$ factors contribute the $y$ is $\binom{n}{k}$, so that term appears $\binom{n}{k}$ times.
4
Summing over all $k$ from $0$ to $n$ gives $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{\,n-k} y^{\,k}$ — and the single $x^{\,n-k}y^{\,k}$ term is read off directly.
Worked Examples
Example 1
Find the coefficient of $x^5$ in the expansion of $(x + 2)^8$.
1
The general term is $\binom{8}{k} x^{\,8-k} 2^{\,k}$; the power of $x$ is $8 - k$.
2
Set $8 - k = 5$, so $k = 3$ selects the $x^5$ term.
Evaluate $\displaystyle\sum_{k=0}^{n} k\dbinom{n}{k}$ in closed form.
1
Start from the binomial identity $(1 + x)^n = \sum_{k=0}^{n}\dbinom{n}{k}x^k$ and differentiate both sides with respect to $x$: $n(1 + x)^{n-1} = \sum_{k=0}^{n} k\dbinom{n}{k}x^{k-1}$.
2
Set $x = 1$: $n\cdot 2^{n-1} = \sum_{k=0}^{n} k\dbinom{n}{k}$.
3
So the sum equals $n\,2^{n-1}$. (Check $n = 4$: $0 + 4 + 12 + 12 + 4 = 32 = 4\cdot 2^{3}$.)
Answer:
$n\,2^{n-1}$.
Going Deeper
Generalization: differentiating, integrating, or substituting roots of unity into $(1 + x)^n$ evaluates whole families of binomial sums; the multinomial theorem extends the expansion to $(x_1 + \cdots + x_m)^n$ with coefficients $\frac{n!}{k_1!\cdots k_m!}$.
Where it appears: AIME 'coefficient of $x^j$' extraction, $\sum \binom{n}{k}$-type identities, probability (the binomial distribution), and counting lattice paths.
Pitfall: a 'minus' or coefficient inside the binomial must travel with its sign and power into EVERY term — in $(2x - 3)^5$ the term is $\binom{5}{k}(2x)^{5-k}(-3)^k$, so the $2^{5-k}$ AND the $(-3)^k$ both matter; dropping the $2^{5-k}$ or the sign on $(-3)^k$ is the classic error.
Spot the Signal
Use it when a power of a binomial appears and you want one specific term or coefficient.
You can describe the hard part as expanding $(x+y)^n$ and reading off the coefficient or term you need, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by writing the general term $\binom{n}{k} x^{n-k} y^k$ and solving for the index $k$ you need.
Name the relation that makes Binomial Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
An off-by-one on the exponent, or forgetting the binomial coefficient entirely.
Applying Binomial Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Write the general term first; then solve for the one index that matters.
Try it on:
Find a specified coefficient or term in a binomial expansion in a contest problem.