Sequences, series, and the binomial theorem
Every formula for listing, summing, or expanding a pattern of terms.
Reach for a sequence formula to find one term directly, without generating everything before it. Move to a series formula once you need a running total instead of a single term — an arithmetic series always finishes, a geometric series only sums infinitely when |r| < 1. The binomial theorem is counting in disguise: its coefficients are exactly a row of combinations, so a shaky expansion usually traces back to a shaky nCr.
Sequences
For an arithmetic sequence with first term a_1 and common difference d, a_n = a_1 + (n-1)d.
For a geometric sequence with first term a_1 and common ratio r \neq 0, a_n = a_1 \, r^{\,n-1}.
Series and summation notation
Term
series
Symbol
\sum
For an arithmetic sequence with first term a_1, common difference d, and nth term a_n, S_n = \frac{n}{2}\left(a_1 + a_n\right) = \frac{n}{2}\bigl(2a_1 + (n-1)d\bigr).
For a geometric sequence with first term a_1 and common ratio r \neq 1, S_n = \sum_{k=1}^{n} a_1 r^{\,k-1} = a_1\,\frac{1 - r^{\,n}}{1 - r}.
For |r| < 1, S = \sum_{k=1}^{\infty} a_1 r^{\,k-1} = \frac{a_1}{1 - r}. The series diverges (has no finite sum) when |r| \ge 1.
The one series formula with a convergence condition attached — it only applies when |r| < 1.
Counting and the binomial theorem
Term
binomial
Term
factorial
The number of ways to choose r items from n distinct items, where order does not matter, is C(n,r) = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}.
Generates the coefficients the binomial theorem expands with.
Show proof
Start from P(n,r), which counts every arrangement of r chosen items. Each unordered group of r items corresponds to many of those arrangements — specifically, every one of the r! ways to reorder that same group is a different permutation but the exact same combination. So P(n,r) overcounts each combination by a factor of r!, and dividing it out gives the true count: C(n,r) = \frac{P(n,r)}{r!} = \frac{n!/(n-r)!}{r!} = \frac{n!}{r!\,(n-r)!}.
For any positive integer n, (a+b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{\,n-k}\, b^{\,k}.
Show proof
Row 1 of Pascal's triangle already matches the claim: (a+b)^1 = a+b = \binom{1}{0}a + \binom{1}{1}b.
Suppose the formula holds for some row n: (a+b)^n = \sum_{k=0}^n \binom{n}{k}a^{n-k}b^k. Multiply both sides by (a+b): (a+b)^{n+1} = \sum_{k=0}^n \binom{n}{k}a^{n+1-k}b^k \ + \ \sum_{k=0}^n \binom{n}{k}a^{n-k}b^{k+1}. Reindex the second sum by k \to k-1 and combine it with the first: the coefficient of a^{n+1-k}b^k in the total is \binom{n}{k}+\binom{n}{k-1}, which is exactly \binom{n+1}{k} by the same row-building identity that produces Pascal's triangle. So (a+b)^{n+1} = \sum_{k=0}^{n+1}\binom{n+1}{k}a^{n+1-k}b^k, the identical formula one row later. That row-to-row step is precisely how each row of Pascal's triangle is generated from the one above it, so the formula holds for every row, starting from row 1.