Sequences, series and finite binomial expansions

What you’ll learn

• Calculate terms and sums of arithmetic and geometric sequences.
• Interpret recurrence and sigma notation.
• Find terms in a positive-integer binomial expansion.

Understand the idea

An arithmetic sequence adds a fixed difference; a geometric sequence multiplies by a fixed ratio. Identify the first term carefully and count transitions: the nth term is reached after $n-1$ steps. A recurrence defines each term from earlier information and can generate increasing, decreasing or periodic behaviour.

For an arithmetic sum, pair first with last: every pair has the same total. For a geometric sum, subtract $rS_n$ from $S_n$ so that the middle terms cancel. The infinite geometric sum exists only when $|r|<1$, because the remaining tail must approach zero.

In a positive-integer binomial expansion, choosing $r$ copies of the second term from $n$ brackets gives the coefficient $\binom nr$. Include both powers and signs in that term. Sigma notation tells you the variable being summed and the first and last indices; it does not mean substitute only the upper limit.

Key formulae

$$u_n=a+(n-1)d,\qquad S_n=\frac n2[2a+(n-1)d]$$

$$u_n=ar^{n-1},\qquad S_n=\frac{a(1-r^n)}{1-r}\quad(r\ne1)$$

$$S_\infty=\frac a{1-r}\quad(|r|<1)$$

$$(a+bx)^n=\sum_{r=0}^n\binom nr a^{n-r}(bx)^r$$

Worked example

A geometric series begins $12,6,3,\ldots$. Find its first five terms' sum and its sum to infinity.

1. The first term is $a=12$ and the ratio is $r=1/2$.
2. Use $S_5=12(1-(1/2)^5)/(1-1/2)=93/4$.
3. Since $|1/2|<1$, the infinite sum exists and equals $12/(1-1/2)$.

Answer: $S_5=93/4=23.25$ and $S_\infty=24$.