Formulae — Σ1, Σn, Σn², Σn³ — proofs: Visual walkthrough

Before line one, here is the entire vocabulary we will use, each earned with a picture as we go. If a word looks scary, wait for its step — it gets a drawing.

Step 1 — What "square a number" looks like

WHAT. $k^{2}$ means a $k \times k$ grid of unit boxes. So $1^{2} = 1$ box, $2^{2} = 4$ boxes, $3^{2} = 9$ boxes.

WHY. Our whole sum $\sum k^{2}$ is literally the total number of little boxes in a growing family of square tiles. Seeing it as area makes the "shortcut" believable — there really is a pattern to count, not just symbols to push.

PICTURE. Look at the growing squares: each new term is a bigger solid square of chalk boxes.

Figure — Formulae — Σ1, Σn, Σn², Σn³ — proofs

Step 2 — The one idea: a difference that cancels itself

WHAT. Pick any function $f (k)$ . Look at the chain of differences $(f (2) - f (1)) + (f (3) - f (2)) + \dots + (f (n + 1) - f (n)) .$ Every middle value $f (2), f (3), \dots, f (n)$ appears once with a plus and once with a minus, so it dies. Only the outermost survivors remain: $\sum_{k = 1}^{n} (f (k + 1) - f (k)) = f (n + 1) - f (1) .$

WHY. This is the engine. It turns a long sum into just two numbers — but only if the thing we are summing can be written as a difference $f (k + 1) - f (k)$ . So our job becomes: find an $f$ whose difference secretly contains $k^{2}$ .

PICTURE. Watch the pluses and minuses annihilate down the ladder; only the top and bottom rungs light up.

Here each symbol's job:

$f (k + 1) - f (k)$ — the rung difference we add up.
$f (n + 1)$ — the top survivor (the very last plus).
$f (1)$ — the bottom survivor (the very first minus).

Step 3 — Choose the right $f$ : why $f (k) = k^{3}$

WHAT. We try $f (k) = k^{3}$ and compute its rung difference: $(k + 1)^{3} - k^{3} .$

WHY $k^{3}$ and not something else? We want a $k^{2}$ to fall out. Expanding a cube produces a square term automatically. Cubing is one power above squaring, and taking a difference of consecutive cubes drops the leading $k^{3}$ (it cancels), leaving the $k^{2}$ we hunt for on top. That is the whole reason for the choice — not magic, just "one power up, then difference."

PICTURE. The cube $(k + 1)^{3}$ is a solid block of side $k + 1$ ; carving out the inner $k^{3}$ block leaves an L-shaped shell. The shell's pieces are exactly $3 k^{2} + 3 k + 1$ .

Expanding with the Binomial Theorem (or just multiplying out): $(k + 1)^{3} - k^{3} = three flat slabs 3 k^{2} + three edge bars 3 k + one corner cube 1 .$ Each term is a physical piece of that shell — the three $k \times k$ slabs are the $3 k^{2}$ , the three $k \times 1$ bars are the $3 k$ , the single corner is the $+ 1$ .

Step 4 — Sum both sides and let the left telescope

WHAT. Apply $\sum_{k = 1}^{n}$ to both sides of Step 3: $\sum_{k = 1}^{n} [(k + 1)^{3} - k^{3}] = \sum_{k = 1}^{n} (3 k^{2} + 3 k + 1) .$

WHY. The left side is exactly the telescoping shape from Step 2 with $f (k) = k^{3}$ , so it collapses instantly. The right side we will break apart in the next step. We are trading a hard sum for an easy end minus start.

PICTURE. The ladder of cube-shells stacks; interiors cancel; only $f (n + 1) = (n + 1)^{3}$ at the top and $f (1) = 1^{3} = 1$ at the bottom survive.

$Left = (n + 1)^{3} - 1.$

$(n + 1)^{3}$ — the top survivor, the biggest cube.
$- 1$ — the bottom survivor, since $f (1) = 1^{3} = 1$ .

Step 5 — Split the right side using known sums

WHAT. Break the right side apart: $\sum_{k = 1}^{n} (3 k^{2} + 3 k + 1) = 3 \sum_{k = 1}^{n} k^{2} + 3 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1.$

WHY. A sum of a sum is the sum of the parts, and a constant multiplier slides out front — this is linearity of $\sum$ . Doing this exposes our unknown $\sum k^{2}$ plus two sums we already know, so we can solve for the unknown.

PICTURE. The mixed pile of chalk boxes sorts into three colour-coded bins: the $k^{2}$ bin (what we want), the $k$ bin (known), and the $1$ bin (known).

The two known building blocks (proven on the parent note and via Arithmetic Progressions for the linear one): $\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}, \sum_{k = 1}^{n} 1 = n .$

$\sum k$ — a triangular number (see Triangular Numbers): stack $1, 2, \dots, n$ into a triangle.
$\sum 1$ — just $n$ ones added, giving $n$ .

Step 6 — Set the two sides equal and solve

WHAT. Equate Step 4 and Step 5, then isolate $\sum k^{2}$ : $(n + 1)^{3} - 1 = 3 \sum k^{2} + 3 \cdot \frac{n ( n + 1 )}{2} + n .$

WHY. The left is a plain polynomial in $n$ ; the right holds our target. One rearrangement frees $3 \sum k^{2}$ , then divide by $3$ .

PICTURE. A balance scale: the cube-difference on the left, the three bins on the right; we slide the known bins across to leave $\sum k^{2}$ alone.

Expand the left, $(n + 1)^{3} - 1 = n^{3} + 3 n^{2} + 3 n$ , then move the known terms over: $3 \sum k^{2} = n^{3} + 3 n^{2} + 3 n - \frac{3 n ( n + 1 )}{2} - n .$ Put over a common denominator of $2$ : $3 \sum k^{2} = \frac{2 n ^{3} + 6 n ^{2} + 6 n - ( 3 n ^{2} + 3 n ) - 2 n}{2} = \frac{2 n ^{3} + 3 n ^{2} + n}{2} .$ Factor the top: $2 n^{3} + 3 n^{2} + n = n (2 n^{2} + 3 n + 1) = n (n + 1) (2 n + 1)$ . So $3 \sum k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{2} ⟹ k = 1 \sum n k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

Step 7 — The degenerate and edge cases (never skipped)

WHAT. Check the boundary values so the formula is trustworthy everywhere.

WHY. A formula that only works for "nice" $n$ is a trap. We test the smallest, the empty case, and the growth for huge $n$ .

PICTURE. Three quick chalk panels: $n = 0$ (nothing), $n = 1$ (one box), and the large- $n$ cubic curve the sum hugs.

$n = 1$ : formula gives $\frac{1 \cdot 2 \cdot 3}{6} = 1$ . Direct: $1^{2} = 1$ . ✔
$n = 0$ (empty sum): the sum has no terms, so it is $0$ by convention. Formula: $\frac{0 \cdot 1 \cdot 1}{6} = 0$ . ✔ The empty sum equalling $0$ is why the start $/$ stop bookkeeping stays consistent.
$n = 2$ : formula $\frac{2 \cdot 3 \cdot 5}{6} = 5$ . Direct: $1 + 4 = 5$ . ✔
Large $n$ : the top is dominated by $2 n^{3}$ , so $\sum k^{2} \approx \frac{2 n ^{3}}{6} = \frac{n ^{3}}{3}$ . This matches $\int_{0}^{n} x^{2} d x = \frac{n ^{3}}{3}$ — the sum is a discrete area, the seed of Definite Integrals as Limits of Sums.

The one-picture summary

Everything compressed: choose $f = k^{3}$ → its shell difference is $3 k^{2} + 3 k + 1$ → summing telescopes the left to $(n + 1)^{3} - 1$ → split the right into the known $\sum k, \sum 1$ → solve for the last bin $\sum k^{2}$ .

Recall Feynman retelling — say it to a friend

We wanted to add up all the square tiles $1, 4, 9, \dots, n^{2}$ without adding them one at a time. Trick: think about cubes instead. Take a cube of side $k + 1$ and peel off the smaller cube of side $k$ inside it — the leftover shell is made of three flat squares ( $3 k^{2}$ ), three thin bars ( $3 k$ ), and one tiny corner ( $+ 1$ ). Now stack these shells for every $k$ from $1$ up to $n$ . Because each shell's inside exactly matches the previous shell's outside, almost everything cancels — like a telescope folding shut — and we're left with just the big cube $(n + 1)^{3}$ minus the tiny starting $1$ . That leftover equals three times our square-sum plus stuff we already know (the plain sum $\frac{n ( n + 1 )}{2}$ and $n$ ). Move the known stuff aside, divide by three, and out drops $\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ . We never added a single square directly — we let the cubes do the counting.

Recall

Which

f (k)

do we telescope for

\sum k^{2}

, and why?

f (k) = k^{3}

, because

(k + 1)^{3} - k^{3} = 3 k^{2} + 3 k + 1

hands us a

k^{2}

term.

What does the left side collapse to after summing

(k + 1)^{3} - k^{3}

(n + 1)^{3} - 1

State

\sum_{k = 1}^{n} k^{2}

\frac{n ( n + 1 ) ( 2 n + 1 )}{6}

Check the formula at

n = 2

\frac{2 \cdot 3 \cdot 5}{6} = 5 = 1 + 4

. ✔

Why is the empty sum (

n = 0

) equal to

0

here?

No terms are added; formula gives

\frac{0 \cdot 1 \cdot 1}{6} = 0

, consistent.

Large-

n

size of

\sum k^{2}

\approx \frac{n ^{3}}{3}

, matching

\int_{0}^{n} x^{2} d x

Connections

Telescoping Series — the collapse in Steps 2 and 4 is exactly this trick.
Binomial Theorem — expands $(k + 1)^{3}$ in Step 3.
Arithmetic Progressions — supplies $\sum k = \frac{n ( n + 1 )}{2}$ used in Step 5.
Triangular Numbers — the shape of $\sum k$ we borrow.
Definite Integrals as Limits of Sums — the large- $n$ cubic match in Step 7.
Mathematical Induction — an alternative way to verify this same formula.
Hinglish version of the parent note