Arithmetic Series: Free Response

Series A gives both the first and last term, so reach for the form built for exactly that.

$S_{23} = \frac{23(9 + 141)}{2} = \frac{23 \times 150}{2}.$

Since $150 \div 2 = 75$,

$S_{23} = 23 \times 75 = 1725.$

Series B gives the first term and the common difference but not the last term, so use the form built from $a_1$ and $d$.

$S_{18} = \frac{18}{2}\big(2(7) + (18-1)(4)\big) = 9\big(14 + 68\big).$

Adding inside the parentheses first,

$S_{18} = 9 \times 82 = 738.$

Substituting the explicit-term formula $a_n = a_1 + (n-1)d$ into $S_n = \frac{n(a_1+a_n)}{2}$ gives

$S_n = \frac{n\big(a_1 + a_1 + (n-1)d\big)}{2} = \frac{n}{2}\big(2a_1 + (n-1)d\big),$

so the $d$-based form is really the first formula in disguise, with the unknown last term replaced by its own formula. That is exactly why it needs $a_1$, $d$, and $n$, but never $a_n$ itself. If Series B's last term had been given directly instead of its common difference, the first form $S_n = \frac{n(a_1+a_n)}{2}$ would skip that substitution and go straight to the answer.

Substitute each whole-number value of the index from the lower limit to the upper limit into the summand, then add the results.

$\sum_{k=2}^{6}(3k-2) = (3\cdot2-2)+(3\cdot3-2)+(3\cdot4-2)+(3\cdot5-2)+(3\cdot6-2).$

That is $4+7+10+13+16$, so

$\sum_{k=2}^{6}(3k-2) = 50.$

Every sigma sum has four pieces: the index (the counting letter), the lower limit (where it starts), the upper limit (where it stops), and the summand (what is being added). Reading $\sum_{k=2}^{6}(3k-2)$ against that pattern, the index is $k$, the lower limit is $2$, the upper limit is $6$, and the summand is $3k-2$.

$\sum_{k=2}^{6}(3k-2).$

Replacing the index letter everywhere it appears, both in the limits and in the summand, gives

$\sum_{i=2}^{6}(3i-2),$

which expands to the very same five terms, in the very same order, giving the same total you already found in part A. The letter used for the index is never part of the final expanded sum; it is only a placeholder that steps through the whole numbers from the lower limit to the upper limit and then disappears once the substitution is done. Because of that, the index is called a dummy variable, and swapping $k$ for $i$, or any other unused letter, changes nothing about the value of the sum.

The daily counts form an arithmetic sequence with first term $a_1=14$ and common difference $d=5$, since each day adds a fixed $5$ bins over the day before. The total over the first $18$ days is the arithmetic series $S_{18}$, and only $a_1$, $d$, and $n=18$ are known, so use the $d$-based form.

$S_{18} = \frac{18}{2}\big(2(14) + (18-1)(5)\big) = 9\big(28 + 85\big).$

Adding inside the parentheses,

$S_{18} = 9 \times 113 = 1017 \text{ bins}.$

This range is its own arithmetic series, with its own first term, last term, and count. There are $15-6+1=10$ days in the range, because both the starting and ending day are counted. The end terms of the range are

$a_6 = 14 + 5(6-1) = 39, \qquad a_{15} = 14 + 5(15-1) = 84.$

Treat the range as a series of $10$ terms running from $39$ to $84$:

$S = \frac{10(39+84)}{2} = 5 \times 123 = 615 \text{ bins}.$

Subtracting the day numbers, $n-m$, counts only the gap between the two days, not the days themselves. Both the starting day $m$ and the ending day $n$ belong to the range and must be counted, so one more day than the gap is actually included. Adding that missing day back in turns $n-m$ into $n-m+1$.

$n - m + 1.$

A small case makes it concrete: days $1$ through $2$ are two days, both counted, even though $2-1=1$.

The first odd number is $1$, so $a_1=1$. The odd numbers increase by $2$ each time, and the $k$th odd number is $2k-1$, so the $n$th term is $a_n=2n-1$.

$a_1 = 1, \qquad a_n = 2n - 1.$

Substituting the values from part A into the sum formula,

$S_n = \frac{n\big(1+(2n-1)\big)}{2}.$

Inside the parentheses, $1+(2n-1)=2n$, so

$S_n = \frac{n(2n)}{2} = \frac{2n^2}{2} = n^2.$

Every step used only the general expressions for $a_1$ and $a_n$, never a specific number for $n$.

Testing the identity at, say, $n=1,2,3$ only shows that it happens to hold for those three values; nothing about that check rules out failure at $n=4$ or beyond. The algebraic argument in part B is different: $n$ was never replaced by a specific number anywhere in the derivation. Every step, from substituting $a_1=1$ and $a_n=2n-1$ to the final simplification, is valid no matter what positive integer $n$ stands for, so the same identity holds for the whole family at once.

$S_1 = 1^2, \quad S_2 = 2^2, \quad S_3 = 3^2, \ \ldots, \ S_n = n^2.$

A finite list of checks can never reach every one of these; a general algebraic argument already covers all of them simultaneously.

Use the explicit formula for a single term.

$a_9 = 8 + (9-1)(6) = 8 + 48 = 56.$

Use a sum formula with $n=9$. Since both $a_1$ and $d$ are known, the $d$-based form works directly.

$S_9 = \frac{9}{2}\big(2(8) + (9-1)(6)\big) = \frac{9}{2}\big(16+48\big).$

Adding inside the parentheses,

$S_9 = \frac{9}{2} \times 64 = 9 \times 32 = 288.$

A term $a_n$ is a single entry of the sequence: it reports the value at one specific position and says nothing about any of the other terms. A sum $S_n$ is a running total: it adds every entry from the first position through the $n$th, so it depends on all of them at once, not just the last one.

$a_n = a_1+(n-1)d, \qquad S_n = a_1+a_2+\cdots+a_n.$

A word problem asking for the amount at one specific position, occurrence, or moment is asking for $a_n$; one asking for a total, a running amount, or how much 'in all' or 'altogether' is asking for $S_n$.