A sequence is a list of numbers; a series is what you get when you add them up. The surprise of this lesson is that an infinite list can still add to a finite number — and that there's a shortcut for raising \( (a+b) \) to any power without ever multiplying it out.
From sequence to series: sigma notation
You already know a sequence: \( 1, 2, 3, 4, \dots \) is just a list, written in order. A series is the running total — what happens when you stop listing and start adding. Adding the first four whole numbers gives the series \( 1 + 2 + 3 + 4 = 10 \).
Writing out long sums by hand is tedious, so mathematicians invented a compact shorthand called sigma notation. The capital Greek letter sigma, \( \Sigma \), is an instruction that says "add these up":
\[ \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n. \]Read it left to right: the \( \Sigma \) says add these up, the \( k = 1 \) underneath is where the counter starts, the \( n \) on top is where it stops, and \( a_k \) is the recipe for each term. So \( \sum_{k=1}^{4} k \) means "let \( k \) run from 1 to 4 and add up every \( k \)" — which is exactly \( 1 + 2 + 3 + 4 = 10 \).
Arithmetic and geometric sums
Two families of series come up again and again, and both have a tidy closed-form sum — a formula that hands you the total without making you add term by term.
An arithmetic series is one where each term differs from the last by a constant difference \( d \), like \( 2 + 5 + 8 + 11 + \cdots \). Its sum has a beautifully simple form:
\[ S_n = \frac{n}{2}\,(a_1 + a_n). \]The trick behind it is Gauss's: pair the first term with the last, the second with the second-to-last, and so on. Every pair adds to the same total \( a_1 + a_n \), and there are \( \tfrac{n}{2} \) such pairs. Legend has it a young Gauss summed \( 1 \) through \( 100 \) in seconds this way.
A geometric series is one where each term is a constant ratio \( r \) times the one before, like \( 3 + 6 + 12 + 24 + \cdots \) (here \( r = 2 \)). Its sum is
\[ S_n = a_1\,\frac{1 - r^n}{1 - r}. \]That single formula collapses any finite geometric run into one division — no matter how many terms.
When infinity is finite
Now the surprise. What if a geometric series never stops — infinitely many terms? Surely an endless sum must blow up to infinity. And it does, unless the ratio is small enough. If \( |r| < 1 \), each term is a shrinking fraction of the one before, so the terms fade away so fast that even infinitely many of them add to a finite total:
\[ S = \frac{a_1}{1 - r}, \qquad |r| < 1. \]This is one of the most beautiful facts in mathematics: an unending sum landing on a single, definite number. Watch the bars below stack toward the dashed limit line. Each new bar is shorter than the last, and the tower climbs but never crosses the line.
Each block is \( r \) times the width of the one before, so the running total \( S_n \) creeps toward \( \dfrac{a_1}{1-r} \) but never overshoots it — add terms and watch the gap shrink, never quite closing.
The binomial theorem
Switch gears to a different kind of shortcut. Expanding \( (a+b)^n \) by hand is brutal for large \( n \): \( (a+b)^2 \) is easy, \( (a+b)^3 \) is a chore, and \( (a+b)^7 \) is a nightmare of bookkeeping. The binomial theorem does the whole expansion in one line:
\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k} b^{k}, \]where the binomial coefficient — the number multiplying each term — is
\[ \dbinom{n}{k} = \frac{n!}{k!\,(n-k)!}. \]The pattern is worth pausing on. As \( k \) climbs from \( 0 \) to \( n \), the power of \( a \) falls while the power of \( b \) rises, and their exponents always sum to \( n \). The coefficient \( \binom{n}{k} \) counts how many ways you can pick \( k \) of the \( b \)'s out of the \( n \) brackets — which is exactly why it shows up.
Pascal's triangle
Here is the gift: you rarely need the factorial formula at all, because those coefficients are exactly the rows of Pascal's triangle. Start with a \( 1 \) at the top, and build each new number as the sum of the two above it. Row \( n \) (counting from \( 0 \)) gives the coefficients of \( (a+b)^n \), in order.
Click any number in the widget below to see its \( \binom{n}{k} \) and the matching expansion light up.
The triangle is symmetric because \( \binom{n}{k} = \binom{n}{n-k} \) — choosing \( k \) things to include is the same as choosing \( n-k \) to leave out.
- The power is \( n = 3 \), so reach for row 3 of Pascal's triangle: \( 1, 3, 3, 1 \). These are the four coefficients.
- Lay out the terms with \( a = x \) and \( b = 2 \), letting the power of \( x \) fall from \( 3 \) to \( 0 \) while the power of \( 2 \) climbs from \( 0 \) to \( 3 \): \( 1\cdot x^3 + 3\cdot x^2(2) + 3\cdot x(2^2) + 1\cdot(2^3) \).
- Now simplify each piece: \( 3\cdot x^2(2) = 6x^2 \), \( 3\cdot x(4) = 12x \), and \( 1\cdot 8 = 8 \).
So \( (x+2)^3 = \mathbf{x^3 + 6x^2 + 12x + 8} \) — written down in seconds, with no multiplying-out at all.
- First find the ratio by dividing any term by the one before it: \( r = \dfrac{2}{6} = \dfrac{1}{3} \). The first term is \( a_1 = 6 \).
- Check the convergence condition: \( |r| = \tfrac{1}{3} < 1 \), so the infinite sum is finite and the formula applies.
- Substitute into \( S = \dfrac{a_1}{1 - r} \): \( S = \dfrac{6}{1 - \tfrac{1}{3}} = \dfrac{6}{\,2/3\,} = 9 \).
Infinitely many terms, and they add to exactly \( S = \mathbf{9} \).
Practice
Try each one yourself, then reveal the full solution.
1. Add the whole numbers from 1 to 100.
This is an arithmetic series with first term \( a_1 = 1 \) and last term \( a_{100} = 100 \), so use \( S_n = \tfrac{n}{2}(a_1 + a_n) \) with \( n = 100 \).
\( S = \tfrac{100}{2}(1 + 100) = 50 \cdot 101 = \mathbf{5050} \).
This is the very sum the young Gauss is said to have computed in seconds, by pairing 1 with 100, 2 with 99, and so on — fifty pairs, each adding to 101.
2. Write the expansion of \( (a+b)^4 \).
The power is \( n = 4 \), so use row 4 of Pascal's triangle: \( 1, 4, 6, 4, 1 \).
Pair each coefficient with falling powers of \( a \) and rising powers of \( b \):
\( (a+b)^4 = \mathbf{a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4} \).
3. Find the sum of \( 1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots \) and its 5th term.
This is geometric with first term \( a_1 = 1 \) and ratio \( r = \tfrac{1}{2} \). Since \( |r| < 1 \), the infinite sum converges.
\( S = \dfrac{a_1}{1 - r} = \dfrac{1}{1 - \tfrac{1}{2}} = \mathbf{2} \).
For the 5th term, the \( k \)th term of a geometric sequence is \( a_1 r^{\,k-1} \), so the 5th term is \( \left(\tfrac{1}{2}\right)^{4} = \mathbf{\tfrac{1}{16}} \).