A sequence is an ordered list of terms, and the $n$th term formula calculates any term directly from its position $n$ ($n=1$ for the 1st term, $n=2$ for the 2nd, and so on) — without listing every term that comes before it. Once a formula is known, finding the 100th term takes exactly the same effort as finding the 3rd.

A sequence is linear when the difference between consecutive terms stays exactly the same. That constant difference, $d$, becomes the coefficient of $n$:

For $7, 11, 15, 19, \ldots$: the common difference is $d=4$, so $c = 7-4=3$, giving $u_n = 4n+3$.

Find the first differences between terms; if those aren't constant, find the second differences (the differences of the differences). If the second differences are constant, the sequence is quadratic and its $n$th term contains an $n^2$ term.

This makes sense because $n^2$ itself always has second differences of exactly $2$ — the values $1,4,9,16,25$ (the square numbers) are the simplest possible quadratic sequence, and every other quadratic sequence is built by shifting or scaling this same underlying pattern.

The coefficient of $n^2$ is always half of the constant second difference. Once you have that, compare the sequence term-by-term against a multiple of $n^2$ to spot what still needs to be added.

Find the $n$th term of $3, 6, 11, 18, 27, \ldots$:

1. First differences: $3, 5, 7, 9$. Second differences: $2, 2, 2$ — constant, so quadratic, with coefficient of $n^2$ equal to $2\div2=1$.
2. Compare the sequence to $n^2 = 1, 4, 9, 16, 25$ — every term is exactly $2$ more:

If the second difference had instead been $4$, the coefficient of $n^2$ would be $4\div2=2$, and the comparison in step 2 would be made against $2n^2 = 2,8,18,32,50$ instead.

A cubic sequence is built from $n^3$. Use the first term to find the multiplier $k$ in $u_n = kn^3$, then check it against a second term before trusting the formula.

The sequence $2, 16, 54, 128, 250, \ldots$ is generated by $u_n=kn^3$. Find $k$ and the 6th term.

1. Using $n=1$: $u_1 = k(1)^3 = k = 2$.
2. Check with $n=2$: $u_2 = 2(2)^3 = 16$ ✓ — matches.
3. Formula: $u_n = 2n^3$, so $u_6 = 2(6)^3 = 432$.

Consider the sequence $7, 11, 15, 19, 23, \ldots$

(a) Write down the next two terms. (b) Find the $n$th term. (c) Find the 50th term.

Solution

(a) Common difference $d=4$, so the next terms are $27$ and $31$.

(b) $c = 7-4=3$, so $u_n = 4n+3$.

(c) $u_{50} = 4(50)+3 = 203$.