Sums of consecutive numbers

A run of consecutive numbers is just a list of whole numbers where each one is one more than the last. If the smallest number is $n$ and there are $k$ of them, the run is

Adding these up, the $n$'s contribute $kn$ in total, and the extra bits $0, 1, 2, \ldots, (k-1)$ add up separately using the well-known triangular-number formula. That gives the general rule:

where: $S$ is the sum, $k$ is the number of terms in the run, and $n$ is the smallest (first) term. A second, often quicker way to say the same thing is sum equals number of terms times the average of the first and last term:

Two shortcuts are worth knowing by heart. First, when $k$ is odd, the run has one exact middle number $m$, and the terms above and below it cancel out in pairs, so $S = k \times m$ — no formula needed, just spot the middle. Second, the sum of the first $n$ natural numbers is the special case $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$. This links straight back to parity (odd/even) reasoning from the same chapter: if $k$ is odd, $S = k \times m$ shows the sum is automatically a multiple of $k$. If $k$ is even, there is no single middle term, and you must fall back on the full formula $S = kn + \dfrac{k(k-1)}{2}$ — a common place students slip up. Why bother with algebra when you could just add the numbers by hand? For a five-term run like $14$ to $18$, direct addition is not hard. But the real power shows up when the question is reversed: given a target total, is it reachable as a sum of, say, seven consecutive numbers, and if so, which seven? Trying every possible starting point by guess-and-check could take a while, especially for large totals. Writing $S = kn + \dfrac{k(k-1)}{2}$ turns that search into one linear equation in $n$ — solve it once, and either you get a clean whole number (the run exists, and you can write it out immediately) or you get a fraction or negative value (the run does not exist, proven instantly, with no case-checking at all). This is exactly the "explain, don't just observe" habit the Number Play chapter is built around: a pattern spotted by trial and error becomes a provable rule once it is written algebraically.

Find the sum of five consecutive numbers starting at 14, and check it two ways.

Step 1. List the run: $14, 15, 16, 17, 18$, so $k=5$ terms starting at $n=14$.

Step 2. Since $k=5$ is odd, spot the middle term: the 3rd number is $16$. So $S = k \times m = 5 \times 16 = 80$.

Step 3. Verify with the general formula: $S = kn + \dfrac{k(k-1)}{2} = 5(14) + \dfrac{5 \times 4}{2} = 70 + 10 = 80$. Both methods agree.

Final answer: $S = \mathbf{80}$.

Competitive stretch

A student claims the sum of four consecutive numbers is 90. Check it, and then decide whether 91 is possible.

Step 1. Let the four numbers be $n, n+1, n+2, n+3$; their sum is $S = 4n + (0+1+2+3) = 4n+6$.

Step 2. Set $4n+6=90$, so $4n=84$ and $n=21$ — a whole number, so it works: $21+22+23+24=90$. Yes, 90 is achievable.

Step 3. Try the same equation for $91$: $4n+6=91 \Rightarrow 4n=85 \Rightarrow n=21.25$, not a whole number, so 91 is impossible. In fact $4n+6$ is always even, so no sum of four consecutive numbers can ever be odd — a parity rule that instantly rules out every odd target.

• Forgetting that an even count of terms ($k$ even) has no single middle number — students wrongly pick just one of the two central numbers instead of using the full formula.
• Assuming every whole number can be written as a sum of consecutive numbers; numbers that are pure powers of $2$ (like $16$ or $32$) cannot be written as a sum of two or more consecutive positive integers.
• Confusing "consecutive numbers" (gap of $1$) with "consecutive even" or "consecutive odd" numbers (gap of $2$), which changes the formula.
• Solving for $n$ but forgetting to check it is a whole (and, if required, positive) number — a fractional or negative $n$ means the target sum cannot be built that way.
• Applying the odd-$k$ middle-term shortcut $S = k \times m$ when $k$ is actually even — always count the terms first before deciding which method to use.
• Mis-stating the offset sum in the derivation as $1+2+\cdots+k$ instead of $0+1+2+\cdots+(k-1)$ — the run starts at offset $0$ (the first term itself), so the last offset is $k-1$, not $k$.

Ask the tutor: Want to see why powers of two can never be written as a sum of consecutive positive integers, or try setting up the equation for a trickier target sum yourself?