Formulae — Σ1, Σn, Σn², Σn³ — proofs: Worked examples

The scenario matrix

Every Σ-problem you will ever meet is one of these cells. The final column names the example that clears it. The flow-map below the table shows the decision path — how you route any incoming problem to its cell.

#	Case class	What makes it tricky	Cleared by
A	Plain single sum, start at $k = 1$	none — pure formula recall	Example 1
B	Linear combination (mixed powers)	must split by linearity	Example 2
C	Shifted start ( $k = a$ , $a > 1$ )	formulas assume start $= 1$ ; subtract the head	Example 3
D	Degenerate / boundary $n = 0, 1$	does the formula still behave?	Example 4
E	Reindex a "weird" sum into standard form	terms not written as $k^{p}$	Example 5
F	Dot-counting / figurate (triangular stack) — NOT a geometric series $r^{k}$	translate a picture into $\sum k$	Example 6
G	Limiting behaviour ( $n \to \infty$ , ratio)	the $\sim n^{p + 1} / (p + 1)$ rule	Example 7
H	Real-world word problem (units!)	model first, then sum	Example 8
I	Exam twist: solve for $n$ (equation)	run the formula backwards	Example 9
J	Steel-man trap: illegal algebra	catch a fake identity	Example 10

Ten examples, ten cells. Let's go.

Cell A — plain recall

Cell B — linear combination

Cell C — shifted start

Cell D — degenerate / boundary

Worked example Example 4 — Do the formulas survive

$n = 0$ and $n = 1$ ? Forecast: an empty sum should be $0$ ; a one-term sum should equal its single term.

Test $n = 0$ (empty sum). By the convention above $\sum_{k = 1}^{0} (\dots) = 0$ (no terms to add). Formula $\sum k = \frac{n ( n + 1 )}{2} = \frac{0 \cdot 1}{2} = 0$ ✔. $\sum k^{2} = \frac{0 \cdot 1 \cdot 1}{6} = 0$ ✔. $\sum k^{3} = [0]^{2} = 0$ ✔. Why this step? A good closed form must return $0$ for the empty case, or it's wrong outside its "tested" range.
Test $n = 1$ (single term). The sum is just the $k = 1$ term. $\sum k = \frac{1 \cdot 2}{2} = 1$ ✔ (term is $1$ ). $\sum k^{2} = \frac{1 \cdot 2 \cdot 3}{6} = 1$ ✔ (term is $1^{2}$ ). $\sum k^{3} = [\frac{1 \cdot 2}{2}]^{2} = 1$ ✔ (term is $1^{3}$ ). Why this step? Boundary $n = 1$ is the smallest non-trivial input; if it fails, the formula is broken.

Answer: all four formulas pass both boundary tests. The plot below shows how each running total climbs from $n = 0$ — read the legend to tell the four curves apart.

Verify: $n = 0 \Rightarrow 0$ and $n = 1 \Rightarrow$ the single term, for every formula ✔.

Cell E — reindex a disguised sum

Cell F — dot-counting / figurate (not a geometric series!)

Worked example Example 6 — How many dots in a triangular stack of

$n$ rows? Forecast: rows have $1, 2, 3, \dots, n$ dots — this is $\sum k$ . Guess $\frac{n ( n + 1 )}{2}$ .

Translate the picture into a sum. Row $k$ holds exactly $k$ dots, so the total is $\sum_{k = 1}^{n} k$ . Why this step? The whole skill is turning a shape into a Σ; see Triangular Numbers.
Apply the formula. $\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$ .
Plug $n = 6$ (the drawn stack): $\frac{6 \cdot 7}{2} = 21$ . Why this step? A concrete count we can literally verify by eye in the figure.

Answer: $\frac{n ( n + 1 )}{2}$ ; for $n = 6$ , that's $21$ dots. Verify: count rows $1 + 2 + 3 + 4 + 5 + 6 = 21$ ✔ (matches the figure's dots).

Cell G — limiting behaviour

Worked example Example 7 — What does

$\frac{\sum _{k = 1}^{n} k ^{2}}{n ^{3}}$ approach as $n \to \infty$ ? Forecast: the parent note said $\sum k^{p} \sim \frac{n ^{p + 1}}{p + 1}$ . For $p = 2$ that's $\frac{n ^{3}}{3}$ , so the ratio should tend to $\frac{1}{3}$ .

Write the exact ratio. $\frac{1}{n ^{3}} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{( n + 1 ) ( 2 n + 1 )}{6 n ^{2}} .$ Why this step? Divide the closed form by $n^{3}$ so the leading powers can compete.
Expand the top. $(n + 1) (2 n + 1) = 2 n^{2} + 3 n + 1$ , giving $\frac{2 n ^{2} + 3 n + 1}{6 n ^{2}}$ . Why this step? To see which term dominates as $n$ grows huge.
Divide every term by $n^{2}$ . $\frac{2 + 3/ n + 1/ n ^{2}}{6}$ . As $n \to \infty$ , the $3/ n$ and $1/ n^{2}$ vanish. Why this step? Anything with $n$ in the denominator shrinks to $0$ for large $n$ .
Take the limit. $\frac{2}{6} = \frac{1}{3}$ .

Answer: the ratio $\to \frac{1}{3}$ . This is exactly $\int_{0}^{1} x^{2} d x$ — see Definite Integrals as Limits of Sums. Verify (numeric, $n = 1000$ ): $\frac{( 1001 ) ( 2001 )}{6 \cdot 1 0 ^{6}} \approx 0.3338$ , close to $0.3 \overline{3}$ ✔.

Cell H — real-world word problem

Cell I — exam twist (solve backwards)

Cell J — steel-man the trap

Worked example Example 10 — Is

$k = 1 \sum n k^{2} = (k = 1 \sum n k)^{?}$ ever clean like the cube case? Statement: A student, dazzled by $\sum k^{3} = (\sum k)^{2}$ , claims $\sum k^{2} = (\sum k)^{3/2}$ . Test it. Forecast: the cube identity is a lucky coincidence for $p = 3$ only — expect this to fail.

Take the smallest real test, $n = 3$ . $\sum k^{2} = 1 + 4 + 9 = 14$ . $\sum k = 1 + 2 + 3 = 6$ . Why this step? One counterexample kills a universal claim — pick a small $n$ where you can compute both sides exactly.
Evaluate the claimed right side. $(\sum k)^{3/2} = 6^{3/2} = 216 \approx 14.70$ . Why this step? Compute the candidate and compare.
Compare. $14 \neq = 14.70$ . The identity is false. Why this step? Even the closest-looking guess breaks; the cube result does not generalise (parent note, Slip 1).

Answer: no such clean power law for $\sum k^{2}$ — always derive, never pattern-match. Verify: $\sum k^{2} = 14$ but $6^{3/2} \approx 14.697 \neq = 14$ ✔ (mismatch confirmed).

Recall Scenario checklist — did we cover every cell?

Plain recall (A) ::: Example 1 Mixed powers / linearity (B) ::: Example 2 Shifted start $k = a$ (C) ::: Example 3, using $\sum_{a}^{b} = \sum_{1}^{b} - \sum_{1}^{a - 1}$ Degenerate $n = 0, 1$ (D) ::: Example 4 — formulas return $0$ and the single term Reindex a disguised sum (E) ::: Example 5 — odd numbers sum to $n^{2}$ Dot-counting / figurate (F) ::: Example 6 — triangular stack $= \frac{n ( n + 1 )}{2}$ Limit / ratio (G) ::: Example 7 — ratio $\to \frac{1}{3}$ Word problem with units (H) ::: Example 8 — $624$ seats Solve for $n$ backwards (I) ::: Example 9 — $n = 12$ Trap / false identity (J) ::: Example 10 — $\sum k^{2} \neq = (\sum k)^{3/2}$ Reversed / empty bounds ::: any $\sum$ with upper $<$ lower is defined to be $0$

Connections

Parent: the four proofs — the formulas drilled here.
Arithmetic Progressions — Example 8's seating is an AP; $\sum k$ is the simplest AP sum.
Triangular Numbers — Example 6's dot stack.
Definite Integrals as Limits of Sums — Example 7's limit is a Riemann sum.
Telescoping Series — the engine behind all four closed forms.
Mathematical Induction — verify any of these by induction instead.
Binomial Theorem — expands $(2 k - 1)$ -style and $(k + 1)^{p}$ terms.