Honors Algebra II Unit 9 Review: Sequences and Series

Mathematical induction proves a statement $P(n)$ is true for every natural number $n$ without checking infinitely many cases. You prove two things. The base case shows $P(1)$ is true. The inductive step shows that if $P(k)$ is true for some arbitrary $k$, then $P(k+1)$ must also be true. The assumption that $P(k)$ holds is called the inductive hypothesis.

Think of a line of dominoes. The base case knocks over the first one, and the inductive step guarantees that every falling domino knocks over the next.

A lot of people worry that assuming $P(k)$ means assuming what you're trying to prove. It doesn't. You never claim $P(k)$ is true by itself. You only prove the link "if $P(k)$, then $P(k+1)$," and the base case supplies the first true statement.

Worked example. Prove $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$. For the base case, $n = 1$ gives a left side of 1 and a right side of $\frac{1 \cdot 2}{2} = 1$, so they match. Now assume $1 + 2 + \cdots + k = \frac{k(k+1)}{2}$. Add the next term, $k + 1$, to both sides. The right side becomes $\frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$. That is exactly the formula with $n = k + 1$, so by induction the formula holds for all $n \ge 1$.

Arithmetic sequences add the same number each time, and geometric sequences multiply by the same number each time. The added number is the common difference $d$, found by subtracting any term from the next one. The multiplier is the common ratio $r$. You can find it by dividing a term by the one before it when that preceding term is nonzero.

Each type has a recursive formula, which builds a term from the previous one, and an explicit formula, which jumps straight to term $n$. For arithmetic sequences these are $a_n = a_{n-1} + d$ and $a_n = a_1 + (n-1)d$. For geometric sequences they are $a_n = r \cdot a_{n-1}$ and $a_n = a_1 r^{n-1}$. For $r = 0$, use $a_1$ as the starting term and the explicit formula for $n \ge 2$. The $n - 1$ appears because the first term has had zero steps applied to it.

Arithmetic sequences follow linear rules with slope $d$. Geometric sequences with $r > 0$ follow exponential rules with base $r$.

Worked example. Find the 8th term of $5, 15, 45, \ldots$ and of $5, 12, 19, \ldots$. The first has $r = \frac{15}{5} = 3$, so $a_8 = 5 \cdot 3^{7} = 5 \cdot 2187 = 10935$. The second has $d = 12 - 5 = 7$, so $a_8 = 5 + 7 \cdot 7 = 54$.

Watch even roots when solving for $r$. If $a_1 = 2$ and $a_3 = 18$, then $r^2 = 9$, so $r = 3$ or $r = -3$, and both are valid unless the problem rules one out.

A series is the sum of the terms of a sequence. Sigma notation writes that sum compactly. In $\sum_{i=1}^{5} (3i - 1)$, the index $i$ runs from the lower limit 1 to the upper limit 5, and you add the general term for each value, so the sum is $2 + 5 + 8 + 11 + 14 = 40$.

Count terms carefully. A sum from $i = m$ to $i = n$ has $n - m + 1$ terms, so $\sum_{i=3}^{10}$ has 8 terms, not 7.

An arithmetic series uses $S_n = \frac{n}{2}(a_1 + a_n)$, which is the number of terms times the average of the first and last terms. A finite geometric series uses $S_n = \frac{a_1(1 - r^n)}{1 - r}$ when $r \ne 1$. If $|r| < 1$, an infinite geometric series has the sum $S = \frac{a_1}{1 - r}$. If $|r| \ge 1$ and $a_1 \ne 0$, the terms do not approach zero, so the series diverges. If $a_1 = 0$, every term is zero and the sum is zero for any $r$.

Worked example. Find $\sum_{n=1}^{\infty} 12\left(\frac{1}{3}\right)^{n-1}$. Here $a_1 = 12$ and $r = \frac{1}{3}$. Since $|r| < 1$, the sum is $\frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 18$. The first four partial sums are approximately $12, 16, 17.33, 17.78$. They are consistent with the sum of 18.