The nth Term from the Sum

Two sums, one term apart

$S_{n}$ is the sum of the first n terms of a sequence: $S_{n} = u_{1} + u_{2} + \dots + u_{n}$. The sum of the first n − 1 terms, $S_{n-1}$, is the same sum without its last term.

So subtracting one from the other cancels every term but the last: $u_{n} = S_{n} - S_{n-1}$. With four terms, $S_{4} - S_{3} = u_{4}$.

From a formula for $S_{n}$

Suppose $S_{n} = n^{2} + 2n$. To get $S_{n-1}$, replace every n with n − 1: $S_{n-1} = \left(n - 1\right)^{2} + 2\left(n - 1\right) = n^{2} - 2n + 1 + 2n - 2 = n^{2} - 1$.

Then $u_{n} = \left(n^{2} + 2n\right) - \left(n^{2} - 1\right) = 2n + 1$. Check it by adding terms: the first four are 3, 5, 7 and 9, which add to 24, and $S_{4} = 16 + 8 = 24$.

The terms go up by 2 each time, so this sequence is arithmetic. That always happens when $S_{n} =$ an² + bn: then $u_{n} = \left(2n - 1\right)a + b$, which goes up by 2a each step.

The first term

$u_{n} = S_{n} - S_{n-1}$ needs a sum of n − 1 terms, so it only makes sense for $n \ge 2$. The first term is found on its own: $u_{1}$ is the first term, which is the sum of one term, so $u_{1} = S_{1}$.

Usually the formula still gives the right $u_{1}$, but not always. With $S_{n} = n^{2} + 2n + 1$ the constant 1 cancels in $S_{n} - S_{n-1}$, so $u_{n} = 2n + 1$ for $n \ge 2$. But $u_{1} = S_{1} = 1 + 2 + 1 = 4$, not 3. The sequence is 4, 5, 7, 9, …, and it is not arithmetic, because its first step is 1.

A geometric example

Take $S_{n} = 3^{n} - 1$. Then $S_{n-1} = 3^{n-1} - 1$, and $u_{n} = 3^{n} - 3^{n-1} = 3 \times 3^{n-1} - 3^{n-1} = 2 \times 3^{n-1}$ for $n \ge 2$.

Check the first term separately: $u_{1} = S_{1} = 3 - 1 = 2$, and $2 \times 3^{0} = 2$ as well, so the formula holds for every n. The terms 2, 6, 18, … form a geometric sequence with first term 2 and ratio 3.

The usual slips

Differentiating $S_{n}$: n counts terms, and the change from one sum to the next is a subtraction, $S_{n} - S_{n-1}$, not a derivative.

Giving $S_{n-1}$, or $S_{k}$ itself, as the term: both are sums of many terms.

Using the formula for $u_{1}$ without checking it against $S_{1}$, which goes wrong whenever $S_{n}$ has a constant term that is not 0.