Sequences and Notation: Practice

Complete the square to see where the rule bottoms out.

$a_n = n^2 - 6n + 12 = (n - 3)^2 + 3$

The square $(n-3)^2$ is never negative, so $a_n \ge 3$, with equality exactly when $n = 3$. Since $3$ is a positive integer it is a legal index, so the smallest term really is attained: $a_3 = 3$. (Checking by hand, the terms open $7,\ 4,\ 3,\ 4,\ 7, \ldots$)

Both limits are included, so subtracting them counts the steps between them and you need one more term than that.

$17 - 4 + 1 = 14$

The index runs $4, 5, 6, \ldots, 17$, which is $14$ values. Answering $13$ is the classic off-by-one.

The four terms are $3$ times $1, 2, 3, 4$, so the summand is $3k$ and the index has to run from $1$ to $4$.

$\sum_{k=1}^{4} 3k = 3 + 6 + 9 + 12 = 30$

Checking the others: $\sum_{k=1}^{3} 3k = 18$ stops one term early, $\sum_{k=2}^{4} 3k = 27$ starts one term late, and $\sum_{k=1}^{4} 3^k = 3 + 9 + 27 + 81 = 120$ puts the index in the exponent.

Climb from the seed, applying the whole rule (multiply by $3$, then subtract $1$) at every step.

$a_2 = 3(2) - 1 = 5, \qquad a_3 = 3(5) - 1 = 14$

One more rung gives the answer.

$a_4 = 3(14) - 1 = 41$

Stopping at $a_3 = 14$ is the common slip here.

The index runs from $2$ to $5$, so there are $5 - 2 + 1 = 4$ terms. Evaluate $k^2 - k$ at each index.

$(4 - 2) + (9 - 3) + (16 - 4) + (25 - 5) = 2 + 6 + 12 + 20 = 40$

The value $54$ is what you get by summing only $k^2$, and $14$ is what you get by summing only $k$.

The factor $(-1)^k$ is $-1$ at odd indices and $+1$ at even ones, so the signs alternate starting with a minus.

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

Dropping the sign factor would give $10$, and starting with a plus instead of a minus would give $-2$.

The stated rule pins the sequence down without any guesswork: start at $4$ and add $3$ each time, giving $4,\ 7,\ 10,\ 13$. Now test each candidate against those terms, since an explicit rule has to match at every index, not just the first.

$a_1 = 3(1) + 1 = 4, \qquad a_2 = 3(2) + 1 = 7, \qquad a_3 = 3(3) + 1 = 10$

The rule $a_n = 4n$ gets the first term right and then gives $8$ instead of $7$; the rule $a_n = n + 3$ gives $4, 5, 6$; and the rule $a_n = 4n - 1$ gives $3, 7, 11$. Only $3n + 1$ matches all the way along.

The graph of a sequence is the set of points $(n, a_n)$, and only positive integers are legal inputs.

$(1, 2), \quad (2, 5), \quad (3, 10), \quad (4, 17), \ldots$

There is no point above $n = 2.5$, because $a_{2.5}$ is undefined, so the graph is a scatter of isolated dots. Joining them into a curve would invent values the sequence does not have.

Both ends are included, so count the indices $5, 6, 7, \ldots, 12$ the same way you count terms in a sum.

$12 - 5 + 1 = 8$

Subtracting alone gives $7$, which counts the steps from $a_5$ to $a_{12}$, not the terms.

Each term is a perfect square, and the numbers being squared run $1$ through $6$, so the summand is $k^2$ with the index from $1$ to $6$.

$\sum_{k=1}^{6} k^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91$

The limits $1$ to $5$ drop the last term ($55$), the limits $2$ to $6$ drop the first ($90$), and cubing gives $441$.

The summand does not mention $k$, so every term is $4$. What the index decides is how many terms there are.

$9 - 3 + 1 = 7 \text{ terms} \quad \Longrightarrow \quad \sum_{k=3}^{9} 4 = 7 \times 4 = 28$

Counting $9 - 3 = 6$ terms instead would give $24$.

A product is zero exactly when one of its factors is zero, so test the indices in order.

$a_1 = (0)(-1) = 0, \qquad a_2 = (1)(0) = 0, \qquad a_3 = (2)(1) = 2$

The first two terms vanish and the third does not, so the smallest such index is $n = 3$. The question asks for the index, not the term $2$.