The Binomial Theorem and Evaluating Binomial Sums

The binomial theorem expands a power of a sum into a weighted sum of monomials, where the weights are the binomial coefficients $\binom{n}{k}=\tfrac{n!}{k!\,(n-k)!}$. Its real konkur value is running it in reverse: when you spot a sum whose terms are $\binom{n}{k}$ times a power, you can fold the whole thing back into a single closed form instead of adding terms one by one.

The theorem

For any reals (or commuting elements) $x,y$ and integer $n\ge 0$,

$(x+y)^{n} \;=\; \sum_{k=0}^{n}\binom{n}{k}\,x^{k}\,y^{\,n-k}.$

The coefficient $\binom{n}{k}$ counts the number of ways to choose the $k$ factors that contribute an $x$ (the rest contribute a $y$), which is exactly why the expansion is combinatorial. The exponents on $x$ and $y$ always sum to $n$, and the coefficients are the entries of row $n$ of Pascal's triangle, obeying $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$.

The one-variable form you actually use

Setting $x=c$ and $y=1$ gives the workhorse identity for collapsing sums:

$\sum_{k=0}^{n}\binom{n}{k}\,c^{k} \;=\; (1+c)^{n}.$

Several standard special cases drop out by choosing $c$:

How to apply

1. Match the shape. Write the summand as $\binom{n}{k}\,(\text{something})^{k}$. Identify $n$ from the binomial coefficient and read off the base $c$ as whatever is raised to the matching power $k$.
2. Check the range. The collapse $\sum_{k=0}^{n}\binom{n}{k}c^{k}=(1+c)^{n}$ needs the index to run over the full range $k=0,\dots,n$. A partial sum does not simplify this way.
3. Substitute and evaluate. Replace the sum by $(1+c)^{n}$ and compute the resulting power.

Mini-example

Evaluate $S=\sum_{k=0}^{6}\binom{6}{k}\,3^{k}$. The summand is $\binom{6}{k}\,3^{k}$, so $n=6$ and $c=3$, and the index runs over the full range. Hence

$S=(1+3)^{6}=4^{6}=4096.$

No term-by-term addition is needed — recognizing the form does all the work.

Exam trick: the instant you see $\sum_{k=0}^{n}\binom{n}{k}c^{k}$, write $(1+c)^{n}$. Confirm the bounds are $0$ to $n$ first; if the sum starts or stops early, peel off the missing terms (e.g. $\sum_{k=1}^{n}=\,2^{n}-1$ for $c=1$) before collapsing.