Sequences and Notation: Free Response

Treat $a_n$ exactly as you would $a(n)$: substitute the index in place of $n$ and simplify.

For the first four terms,

$a_1 = 1^2 + 2(1) - 1 = 2, \qquad a_2 = 2^2 + 2(2) - 1 = 7,$

$a_3 = 3^2 + 2(3) - 1 = 14, \qquad a_4 = 4^2 + 2(4) - 1 = 23.$

Reaching a distant term costs no more than one substitution, because the rule is explicit:

$a_{20} = 20^2 + 2(20) - 1 = 400 + 40 - 1 = 439.$

No earlier term was needed to find $a_{20}$; the rule was fed the index $20$ and returned a value directly.

Read the definition backward from the value. Substituting $n = 3$ into the rule from part A confirms which index is responsible:

$a_3 = 3^2 + 2(3) - 1 = 14.$

So the index that produced $14$ is $3$, not $14$ itself.

The confusion "the index is $14$" mixes up the two roles a sequence keeps separate: the index tells you WHICH term of the list you are looking at, and the term is WHAT that entry is worth. Here the index is the small number $3$, counting a position in the list, while $14$ is the (generally much larger) output the rule produced once it was fed that position. Nothing about the size of a term says anything about the size of the index that produced it.

The reason has nothing to do with this particular formula, so the argument should not appeal to it. A sequence is a function whose domain is fixed in advance:

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

Asking for $a_{2.5}$ is asking the function to accept an input, $2.5$, that was never in its domain to begin with.

This is exactly the situation of an ordinary function evaluated outside where it is defined: the expression is not a hard computation that happens to give a strange answer, it is not a legitimate question about the function at all. Substituting $2.5$ into the formula $n^2 + 2n - 1$ would certainly produce a number, but that number would not be a term of this sequence, because the sequence was never asked to say anything about non-integer inputs.

This rule needs two terms back, so both seeds are used immediately, and each new term uses the two terms directly before it.

At $n = 3$, both seeds are used:

$c_3 = c_2 + c_1 + 1 = 4 + 3 + 1 = 8.$

The next two terms each use the pair just computed:

$c_4 = c_3 + c_2 + 1 = 8 + 4 + 1 = 13, \qquad c_5 = c_4 + c_3 + 1 = 13 + 8 + 1 = 22.$

Each term used the two terms immediately before it, never anything further back.

"Looking back two terms" describes what one application of the rule uses, not how far the rule can reach in total. Applying the rule at $n = 5$ requires $c_4$ and $c_3$, neither of which is a seed:

$c_5 = c_4 + c_3 + 1.$

So before $c_5$ can be produced, $c_3$ has to be built from the two seeds, and then $c_4$ has to be built from $c_3$ and $c_2$. Only once both of those exist does the rule at $n = 5$ have the two ingredients it needs. There is no shortcut that lets the rule reach directly from $c_1$ and $c_2$ to $c_5$: every rung of the ladder between the seeds and the target term has to be climbed.

The pattern from parts A and B generalizes directly. A rule that reaches back $k$ terms cannot be applied until $k$ consecutive terms already exist to feed it, and the only way to have terms with no earlier terms behind them is to state them as seeds.

For $d_n = d_{n-1} + d_{n-2} + d_{n-3}$, the first index the rule can be applied at is $n = 4$, and applying it there uses $d_3$, $d_2$, and $d_1$, all three at once:

$d_4 = d_3 + d_2 + d_1.$

None of those three can be produced by the rule itself, since each of them would need terms with an even smaller index, running past where the sequence starts. So all three, $d_1$, $d_2$, $d_3$, have to be given directly, and $d_4$ is the first term the rule actually builds.

A bare rule fixes only the STEP from one term to the next, so testing the claim means checking whether two different starting choices can both be made to obey it.

Take $e_1 = 1$ and apply the rule three times:

$e_2 = 1 + 3(2) = 7, \qquad e_3 = 7 + 3(3) = 16, \qquad e_4 = 16 + 3(4) = 28.$

Now take $e_1 = 4$ instead, and apply the same rule:

$e_2 = 4 + 3(2) = 10, \qquad e_3 = 10 + 3(3) = 19, \qquad e_4 = 19 + 3(4) = 31.$

Both $1, 7, 16, 28, \ldots$ and $4, 10, 19, 31, \ldots$ satisfy $e_n = e_{n-1} + 3n$ at every step, and yet they are two different sequences that never agree. So the bare rule does not pin down one sequence; it describes an entire family of them, one for every choice of $e_1$.

What has to be shown is that the gap present at the seed never closes, however far the sequences run. Work with the difference $e_n - f_n$ directly.

At $n = 1$, the difference is $e_1 - f_1$ by definition, and this is nonzero by hypothesis.

Now suppose the difference at some index $n - 1$ (with $n \ge 2$) equals $e_1 - f_1$. Both sequences obey the same rule, so each adds the identical quantity $3n$ to move from index $n-1$ to index $n$:

$e_n - f_n = (e_{n-1} + 3n) - (f_{n-1} + 3n) = e_{n-1} - f_{n-1}.$

By the assumption for index $n - 1$, that last quantity is $e_1 - f_1$. So the difference at index $n$ is again $e_1 - f_1$, unchanged from the difference one step earlier.

The difference therefore equals $e_1 - f_1$ at $n = 1$, and each step forces the next index to inherit that exact same value, so it equals $e_1 - f_1$ at every positive integer $n$. Since $e_1 \neq f_1$, this common value is nonzero at every index, so $e_n \neq f_n$ for every $n$. The gap created by the two different seeds never narrows, because the rule adds the same amount to both sequences at every step and an equal addition to two unequal numbers cannot make them equal.

Describe a single term of the sum before describing the whole sum. Each term here is an integer squared, and the integer being squared counts upward by one from term to term: $5, 6, 7, 8, 9$. So the term whose base is $k$ is simply $k^2$, and the limits of the index are fixed by where that base starts and stops:

$5^2 + 6^2 + 7^2 + 8^2 + 9^2 = \sum_{k=5}^{9} k^2.$

Check the two ends before trusting it: at $k = 5$ the summand is $5^2$, and at $k = 9$ it is $9^2$, which are exactly the first and last terms of the original sum.

Count the terms before adding anything, using the rule that both limits are included: the index runs from $5$ to $9$, giving $9 - 5 + 1 = 5$ terms.

Evaluate the summand at each of those five indices,

$5^2 = 25, \quad 6^2 = 36, \quad 7^2 = 49, \quad 8^2 = 64, \quad 9^2 = 81,$

and add the results in order:

$25 + 36 + 49 + 64 + 81 = 255.$

The arithmetic the student performed on each of the four terms they kept is correct; the error is entirely in how many terms they decided to keep.

Subtracting the limits, $9 - 5 = 4$, counts the number of STEPS between $5$ and $9$, not the number of integers in that range. The indices actually used are $5, 6, 7, 8, 9$, five numbers in total, and the student's total of four silently dropped the last one, $k = 9$.

Because both limits of a sigma sum are included, the correct count is

$9 - 5 + 1 = 5,$

and restoring the missing term $9^2 = 81$ gives the correct total:

$174 + 81 = 255.$

The index is not itself a term of the sum; it is an instruction for which values of $a_k$ to add. Whatever letter carries that instruction, the instruction says the same thing: start at $m$, walk up through the integers one at a time, stop at $p$, and add the value of $a$ at each stop:

$\sum_{k=m}^{p} a_k = a_m + a_{m+1} + \cdots + a_p.$

Changing the letter used for that walk, from $k$ to any other symbol not already in use elsewhere in the expression, changes nothing about which integers are walked through or what is evaluated at each one. The list of numbers actually being added, $a_m, a_{m+1}, \ldots, a_p$, is identical either way, and a sum of the same list of numbers has the same total.

That is also why the index vanishes once the sum is worked out: it never named a value in the answer, only a rule for producing the list of values that got added.

A renaming is only safe when the new letter is not already doing another job in the expression, and here it is: $m$ was the upper limit before the student ever touched the index.

After the substitution, the same symbol $m$ is asked to name the fixed upper limit of the sum and, at the same time, the value that walks from $1$ up to that limit one step at a time. Those are two different jobs, a constant and a variable, and one letter cannot honestly do both in the same expression:

$\sum_{m=1}^{m} (3m - m).$

That expression does not describe the original sum, or indeed any sum with a clear meaning at all, since it is unclear at any point which occurrence of $m$ is the fixed limit and which is the walking index.

The fix is to choose a letter not already in use anywhere in the expression, for instance $j$: $\sum_{j=1}^{m}(3j-m)$ says exactly what the original sum said.

This sum is unusual in one useful way: because the summand is always $1$, the value of the sum and the number of terms it has are literally the same quantity, so proving the sum equals $p - m + 1$ is the same as proving the term-count rule directly.

The index $k$ runs through the consecutive integers $m, m+1, m+2, \ldots, p$. Shift every one of them down by $m - 1$: the list becomes

$1, 2, 3, \ldots, p - m + 1,$

and shifting every entry of a list by the same amount changes none of the entries relative to each other and drops none of them, so this new list has exactly as many entries as the original one. The shifted list is plainly the consecutive integers from $1$ to $p - m + 1$, which has $p - m + 1$ entries.

Since the original list of indices has that same count, and the sum adds exactly $1$ once for each index in that list,

$\sum_{k=m}^{p} 1 = \underbrace{1 + 1 + \cdots + 1}_{p-m+1 \text{ times}} = p - m + 1.$

Because this sum's value is, by construction, nothing other than a count of how many indices ran from $m$ to $p$, this single computation proves the general counting rule: any sum whose index runs from $m$ to $p$, whatever its summand, has $p - m + 1$ terms.