MISSION · Algebra
Adding Up a Sequence
How Gauss added 1 to 100 in seconds
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The idea in one sentence
A sequence is a list; a series is that list added up — and both kinds of sequence have a formula that beats adding the terms one at a time.
The trick behind the arithmetic formula
The story goes that Gauss, aged about nine, was told to add the numbers from 1 to 100 and answered almost immediately.
Pair the terms from the outside in: \(1 + 100 = 101\), \(2 + 99 = 101\), \(3 + 98 = 101\). Every pair makes 101, and there are 50 pairs:
\[50 \times 101 = 5050\]That pairing is the whole formula. In general, each pair is (first + last), and there are \(\frac{n}{2}\) of them:
\[S_n = \frac{n}{2}(a_1 + a_n)\]Read it as “the number of terms, times the average of the first and last”. That reading also covers an odd number of terms, where the pairing leaves a middle value — which is exactly the average anyway.
If you know the common difference rather than the last term, substitute \(a_n = a_1 + (n-1)d\):
\[S_n = \frac{n}{2}\left(2a_1 + (n-1)d\right)\]Same formula, different information supplied.
The geometric formula
\[S_n = a_1\frac{1 - r^n}{1 - r} \qquad (r \ne 1)\]The derivation is a neat piece of algebra worth seeing once. Write out \(S\), then write out \(rS\) beneath it shifted along one place, and subtract. Almost every term cancels, leaving \(S - rS = a_1 - a_1r^n\), which rearranges to the formula above.
For \(3 + 6 + 12 + 24 + 48\), with \(a_1 = 3\), \(r = 2\), \(n = 5\):
\[S_5 = 3\frac{1 - 32}{1 - 2} = 3 \times 31 = 93\]And the terms do add to 93.
Infinite geometric series
When \(|r| < 1\) the terms shrink fast enough that the total settles down, and \(r^n\) vanishes as \(n\) grows:
\[S_\infty = \frac{a_1}{1 - r}\]That’s the finite formula with the \(r^n\) term gone. When \(|r| \ge 1\) there is no such total — the sum runs away.
Sigma notation
\[\sum_{k=1}^{5} 2k = 2 + 4 + 6 + 8 + 10 = 30\]The \(\Sigma\) says “add up”. Below it is where the counter starts, above it where it stops, and beside it the rule for each term. It’s a compact instruction, not an operation to be afraid of — expand two or three terms and the pattern is always obvious.
Common mix-ups
- Confusing sequence and series. One lists, the other totals.
- Miscounting the terms. From 5 to 20 inclusive is 16 terms, not 15.
- Using the geometric formula with \(r = 1\), which divides by zero. If \(r = 1\) every term is the same, so the sum is \(n a_1\).
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**Using the infinite formula when $$ r \ge 1$$.**
Check yourself
Find \(2 + 5 + 8 + \cdots + 29\).
Arithmetic with \(d = 3\). The number of terms: \(29 = 2 + 3(n-1)\) gives \(n = 10\). Then \(S = \frac{10}{2}(2 + 29) = 5 \times 31 = 155\).
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