Sequences and Notation: Practice

The definition restricts the inputs of the function to the counting numbers.

$\text{domain} = \{1, 2, 3, 4, \ldots\}$

That restriction is the whole difference between a sequence and a general function. Because the inputs arrive in order, the outputs can be written as a list.

The subscript is the index, which says which term you want, not what the term is worth.

$a_1 = 5, \qquad a_2 = 9, \qquad a_3 = 13$

Count along to the third position: $a_3 = 13$. Answering $3$ would be confusing the index with the term.

A recursive rule has to be climbed one rung at a time, starting from the seed. First use $a_1$ to get $a_2$.

$a_2 = a_1 + 6 = 4 + 6 = 10$

Now feed $a_2$ back into the same rule.

$a_3 = a_2 + 6 = 10 + 6 = 16$

The index $k$ runs from the lower limit $1$ up to the upper limit $4$, and the summand is $k$ itself, so write out each term and add.

$\sum_{k=1}^{4} k = 1 + 2 + 3 + 4 = 10$

Evaluate the summand $2k$ at each index from $1$ to $3$, then add the three results.

$\sum_{k=1}^{3} 2k = 2(1) + 2(2) + 2(3) = 2 + 4 + 6 = 12$

Answering $6$ would be evaluating the summand only at the last index instead of adding all three.

For a sequence the parentheses of function notation are replaced by a subscript, and nothing else changes.

$a_n = a(n) \quad \Longrightarrow \quad a_7 = a(7)$

The subscript is an index, not an exponent and not a multiplier.

The rule says how to get from one term to the next, but it never says where to begin, so it does not single out one sequence.

$a_1 = 1 \Rightarrow 1, 3, 9, 27, \ldots \qquad a_1 = 5 \Rightarrow 5, 15, 45, 135, \ldots$

Both of those obey $a_n = 3a_{n-1}$, and so does one sequence for every other choice of $a_1$. A recursive definition needs its seed.

Substitute $n = 3$ and remember that a negative base raised to an odd power stays negative.

$a_3 = (-1)^3 = -1$

This sequence is $-1, 1, -1, 1, \ldots$, and the factor $(-1)^n$ is the standard way to make terms alternate in sign.

The lower limit is $2$, not $1$, so the index takes only the values $2$, $3$, and $4$.

$\sum_{k=2}^{4} k^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$

Starting at $k = 1$ out of habit would have added the extra term $1^2 = 1$, giving $30$.

The index of summation is the letter written under the sigma, the one that counts upward through the terms.

$\sum_{i=1}^{6} 3i = 3(1) + 3(2) + \cdots + 3(6)$

Here $1$ is the lower limit, $6$ is the upper limit, and $i$ is the index. The $3$ is part of the summand.

The first term is the output at the first index, and the index starts at $n = 1$, not at $n = 0$.

$a_1 = 7(1) - 3 = 4$

Substituting $n = 0$ would give $-3$, but $0$ is not in the domain of this sequence.

Substitute $n = 2$ into both the numerator and the denominator.

$a_2 = \frac{2}{2 + 1} = \frac{2}{3}$

The term $\frac{1}{2}$ is $a_1$, not $a_2$.