Formulae — Σ1, Σn, Σn², Σn³ — proofs: Question bank

Before we start we must pin down the one symbol that runs through every line of this page.

Three ideas about that loop carry the whole page:

Two more symbols appear on this page, so let us earn them before use:

True or false — justify

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

because we are summing the constant

1

False. The constant

1

is added once per term, and there are

n

terms, so the sum is

n

, not

1

. Constant summand

\neq =

constant sum.

\sum_{k = 1}^{n} k^{3} = (\sum_{k = 1}^{n} k)^{2}

, and by the same logic

\sum k^{2} = (\sum k)^{3/2}

False. The cube identity is a genuine coincidence special to

p = 3

; there is no matching clean power-law for

\sum k^{2}

, whose real form is

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

. Never generalise a coincidence.

\sum_{k = 1}^{n} (k + 3) = \sum_{k = 1}^{n} k + 3

False. The

+ 3

is added inside every term, so it contributes

3 n

, not

3

: the correct value is

\frac{n ( n + 1 )}{2} + 3 n

. Only a constant factor pulls out cleanly; a constant added term multiplies by the count.

For large

n

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

grows like

\frac{1}{3} n^{3}

True. The leading term of

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

\frac{2 n ^{3}}{6} = \frac{n ^{3}}{3}

, matching

\int_{0}^{n} x^{2} d x = \frac{n ^{3}}{3}

; the leftover

\frac{n ^{2}}{2} + \frac{n}{6}

is a lower-order truncation term that fades relative to

n^{3}

. See the Riemann-rectangles figure below.

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

is the same as an arithmetic-progression sum.

True. It is the AP with first term

a = 1

, common difference

d = 1

, giving

\frac{n}{2} (2 a + (n - 1) d) = \frac{n ( n + 1 )}{2}

— exactly the Arithmetic Progressions formula. The Gauss-pairing figure below shows the two stacked staircases forming a rectangle.

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

contains a

k^{2}

term, which is why we chose

f (k) = k^{3}

to sum squares.

True. Expanding gives

3 k^{2} + 3 k + 1

; the

k^{2}

appears, so telescoping this difference exposes

\sum k^{2}

once the known

\sum k

and

\sum 1

are subtracted off.

Telescoping only works when the term is already written as a difference

f (k + 1) - f (k)

False (in spirit). We deliberately manufacture the difference by choosing a clever

f

(e.g.

k^{3}

k^{4}

); the term

k^{2}

is not a difference until we build

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

around it. The skill is inventing

f

, not spotting one.

\sum_{k = 0}^{n} k = \sum_{k = 1}^{n} k

True. The

k = 0

term is

0

, so adding it changes nothing. Zero start terms are harmless — but never assume this for other functions (e.g.

k^{0}

\frac{1}{k}

Spot the error

\sum_{k = 1}^{n} 3 k^{2} = (\sum_{k = 1}^{n} 3 k)^{2}

." — find the mistake.

Two errors: you cannot square-out an exponent from a sum, and

3 k^{2} \neq = (3 k)^{2} = 9 k^{2}

. Correct:

\sum 3 k^{2} = 3 \sum k^{2}

— the constant

3

pulls out, the square stays inside.

\sum_{k = 5}^{12} k^{2} = \frac{12 \cdot 13 \cdot 25}{6} = 650

." — what's wrong?

The formula counts from

k = 1

, so this includes the unwanted

k = 1..4

head. Fix: subtract it —

\sum_{1}^{12} - \sum_{1}^{4} = 650 - 30 = 620

"Since

\sum_{k = 1}^{n} (2 k + 1) = (n + 1)^{2} - 1

, we have

\sum k = \frac{( n + 1 ) ^{2} - 1}{2}

." — spot the slip.

The left side is

2 \sum k + \sum 1 = 2 \sum k + n

, not

2 \sum k

. Forgetting the

+ n

loses a term. Correctly

2 \sum k + n = (n + 1)^{2} - 1 \Rightarrow \sum k = \frac{n ( n + 1 )}{2}

\sum_{k = 1}^{n} (k + 1)^{3} - k^{3} = (n + 1)^{3} - 1^{3}

, and

1^{3} = 1

, so it's

(n + 1)^{3} - 1

; but I'll also cancel the

k = 1

lower limit as

f (0)

." — locate the confusion.

Telescoping gives

f (n + 1) - f (1)

where

f (k) = k^{3}

, so the leftover bottom is

f (1) = 1

, not

f (0) = 0

. The surviving lower endpoint is set by the starting index, here

k = 1

Σ

is linear, so

\sum k^{2} = (\sum k) \cdot (\sum k) / n

." — why is this nonsense?

Linearity applies to sums of terms, not to products of whole sums.

\sum k^{2}

is not built from

\sum k

by any algebraic shortcut of this kind; only a direct derivation gives

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

"The

6

4 k^{3} + 6 k^{2} + 4 k + 1

from

(k + 1)^{4} - k^{4}

is a lucky number." — is it?

No — it's the binomial coefficient

(2 4) = 6

, the middle entry of row 4 of Pascal's triangle

1, 4, 6, 4, 1

. See the coefficient-bar figure below; every coefficient in a difference

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

is a Pascal number by the Binomial Theorem.

Why questions

Why do we pick

f (k) = k^{4}

(not

k^{3}

) to derive

\sum k^{3}

Because we need the difference

f (k + 1) - f (k)

to contain a

k^{3}

term. Expanding

(k + 1)^{4}

by Pascal's row

1, 4, 6, 4, 1

and subtracting

k^{4}

leaves

4 k^{3} + 6 k^{2} + 4 k + 1

— the top power drops by exactly one, so choosing

f

one power higher than the target always exposes it. The step-by-step build is in the figure above.

Why does telescoping "collapse" a sum at all?

Each interior value

f (k)

appears once as

+ f (k)

(from the

k - 1

term's

f ((k - 1) + 1)

) and once as

- f (k)

, so it cancels — exactly the paired green/red arrows in figure s02. Only the very first minus-piece

f (1)

and very last plus-piece

f (n + 1)

have no partner.

Why is

\sum k^{p} \approx \frac{n ^{p + 1}}{p + 1}

for large

n

(with

p

a non-negative integer)?

A sum of

n

rectangles of width

1

and height

k^{p}

approximates the area under

y = x^{p}

, and

\int_{0}^{n} x^{p} d x = \frac{n ^{p + 1}}{p + 1}

; the mismatch between step-tops and the smooth curve is a lower-order error. This is the bridge to Definite Integrals as Limits of Sums.

Why is the Binomial Theorem lurking in every one of these proofs?

Each proof expands

(k + 1)^{p}

; the coefficients

1, p, (2 p), \dots

come straight from Binomial Theorem (Pascal's triangle). That's what turns a difference into a polynomial in

k

we can split.

Why can we substitute

\sum k

and

\sum 1

while deriving

\sum k^{2}

, but not use

\sum k^{2}

itself?

We build the formulas in order of power: each new one relies only on the already-proved lower ones. Using the very quantity we're solving for would be circular — it's the unknown.

Why does Gauss's pairing give the same answer as telescoping for

\sum k

Both are valid, independent routes to a true fact. Pairing sums forwards+backwards into

n

columns of

(n + 1)

(figure s04); telescoping cancels a manufactured difference. Different pictures, one truth — a good sign the result is right.

Edge cases

What is

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

when

n = 0

(empty sum)?

0

. An empty sum (no terms) is defined as

0

, and the formula agrees:

\frac{0 \cdot 1}{2} = 0

. The closed form gracefully covers the degenerate case.

What is

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

when

n = 1

1

. Only the

k = 1

term exists; check:

\frac{1 \cdot 2 \cdot 3}{6} = 1

. Always sanity-test a formula at

n = 1

— it's the cheapest error-catcher.

Does

\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}

still hold if

n

is, say, negative or fractional?

The derivation assumes

n

is a positive integer (you can't sum a fractional number of terms). The polynomial on the right still evaluates, but it no longer means "a sum of cubes" — the identity is a statement about positive integers only.

If a sum starts below

1

, e.g.

\sum_{k = - 2}^{4} k

, can we still use the shift trick? Show it numerically.

Yes — split off the head by hand:

\sum_{k = - 2}^{4} k = = - 3 (- 2) + (- 1) + 0 + \sum_{k = 1}^{4} k = - 3 + \frac{4 \cdot 5}{2} = - 3 + 10 = 7

. Direct check:

- 2 - 1 + 0 + 1 + 2 + 3 + 4 = 7

. ✔ The standard formula only owns the

k \geq 1

tail; the negative/zero head is added separately.

What does

\sum_{k = a}^{b}

equal when

a = b

(single term) or

a > b

(backwards)?

a = b

it's just the one term

u_{a}

. If

a > b

the range is empty, so the sum is

0

by convention. Both fall out of

\sum_{1}^{b} - \sum_{1}^{a - 1}

if you trust the algebra — but always eyeball whether the range makes sense.

\sum k^{3} = (\sum k)^{2}

ever equal to

\sum k^{2}

Only in the trivial case

n = 1

, where all four basic sums equal

1

. For any

n \geq 2

the three sums diverge in size, so the coincidence

\sum k^{3} = (\sum k)^{2}

does not spill over to

\sum k^{2}

Connections

Telescoping Series — the cancellation engine behind every "why" here.
Arithmetic Progressions — the true/false item on $\sum k$ as an AP.
Mathematical Induction — the trap-free alternative if you distrust a manufactured $f$ .
Definite Integrals as Limits of Sums — why $\sum k^{p} \approx \frac{n ^{p + 1}}{p + 1}$ .
Binomial Theorem — the expansions hidden in each difference.
Triangular Numbers — the object squared in the cube identity.