Σ is an instruction, not a mystery: run the index from the bottom number to the top, feed each value into the formula, add everything up.
The problem
Σk=14 (2k + 1)
Run k through 1, 2, 3, 4: the terms are 3, 5, 7, 9 — sum 24. C computes only the last term; A drops the +1 from each of the four terms. Small sums are meant to be expanded by hand.
| Piece | Job |
|---|---|
| k = 1 (below) | where the index starts |
| 4 (above) | where it stops — inclusive |
| 2k + 1 (beside) | the recipe each k is fed into |
The index letter is disposable — Σ over k of 2k + 1 and Σ over i of 2i + 1 are the same sum. Only the start, the stop and the recipe matter.
k = 1, 2, 3, 4 → 3, 5, 7, 9
each k through the recipe once
3 + 5 + 7 + 9 = 24
then just add
k = 3 to 7 → 7 − 3 + 1 = 5 terms
both ends count — the classic off-by-one
Σ of a constant: c added n times = cn
no k in the recipe? every term is the same
Σk=1100 k = 100 · 1012 = 5,050
n(n + 1)2 — the arithmetic series in Σ clothing
Σ (2k + 1) = 2·Σk + Σ1
Σ distributes over + and pulls constants out
The Σ symbol changes the packaging, never the mathematics: an arithmetic recipe inside a Σ is an arithmetic series, formulas included.
The stop is the unknown. Σ (k = 1 to n) of (2k + 1) = 99. Find n.
2 · n(n + 1) ÷ 2 + n = n² + 2n
2·Σk + Σ1, with Σ1 = n
n² + 2n − 99 = 0 → (n + 11)(n − 9) = 0
set it equal to 99
n = 9
a count cannot be negative
The same splitting works on letters: if Σ aₖ = 12 and Σ bₖ = 5 over k = 1 to 8, then Σ (aₖ + 4bₖ − 2) = 12 + 4(5) − 2(8) = 16. The constant −2 is added 8 times.
the terms rise by 4
so the recipe is 4k + c
k = 1: 4(1) + c = 7, so c = 3
fit the first term
last term: 4k + 3 = 27, so k = 6
fit the last term
Σ (k = 1 to 6) of (4k + 3)
Test both ends: 4(1) + 3 = 7 and 4(6) + 3 = 27 ✓, and 6 − 1 + 1 = 6 terms ✓.
Σ (k = 0 to 6) of (4k + 3) has the right recipe but begins with an extra 3 — seven terms, not six.
Alternating signs. Find Σ (j = 1 to 7) of j² × (−1)j.
−1 + 4 − 9 + 16 − 25 + 36 − 49
odd j gives a minus
(4 − 1) + (16 − 9) + (36 − 25) − 49
pair them up; j = 7 is left over
3 + 7 + 11 − 49 = −28
the odd term keeps its minus
What is different: (−1)k only sets the sign. Write the signs first, then pair; with an odd number of terms one is left unpaired.
Starting at k = 0. Evaluate Σ (k = 0 to 5) of 2k.
5 − 0 + 1 = 6 terms
k = 0 counts
1 + 2 + 4 + 8 + 16 + 32
20 = 1
= 63
or 26 − 1, a geometric series
What is different: the first term is the recipe at k = 0, which is 1 here, not 0. The recipe is geometric, so for a long sum use the geometric series formula.
Starting the list at 2 gives 62, one term short; counting “5 terms” and stopping at 16 gives 31. Write the k = 0 term first.
Σ (k = 1 to 40) of k = 40 × 41 ÷ 2 = 820
everything up to 40
Σ (k = 1 to 20) of k = 20 × 21 ÷ 2 = 210
remove 1 to 20 — not 1 to 21
820 − 210 = 610
Second route: 40 − 21 + 1 = 20 terms averaging (21 + 40) ÷ 2 = 30.5, and 20 × 30.5 = 610 ✓.
The sum starts at 21, so 21 stays in. Take away 1 to 20 only.
Σ questions test reading as much as adding: evaluate a short sum, pick the Σ that matches a written-out sum, or handle a long sum that starts above 1. Count terms as stop − start + 1, and split long sums into Σk and constants. Ten-second check: write out the first and last terms of your Σ and compare them with the question.
The problem
Σk=37 k²
Five terms (7 − 3 + 1): 9 + 16 + 25 + 36 + 49 = 135. D starts at k = 1, sneaking in the extra 1 + 4, and C runs one term too far, to k = 8 — the start value is part of the instruction, not decoration.
