Summation

We traditionally speak of a sum from i = a to i = b, where b − a is a whole number (0, 1, 2, …); assuming for simplicity that a is an integer (so that b is also), this sum covers every integer i that satisfies the inequality a ≤ i ≤ b, or in other words all of the integers in the interval [a, b]. But in many ways, it's better to think of such a sum as running from i = a to i = b + 1, but with the last item not quite included; that is, the sum covers every integer i that satisfies the inequality a ≤ i < b + 1, or in other words all of the integers in the interval [a, b + 1). Of course, from this perspective, it's not the number b that matters but the number b + 1; if we call this B, then we can write

a≤i<B

for what is normally written as

b
Σ	.
i=a

Note also that it makes perfect sense to have B = a (in other words, b − a = −1); then we are adding up no terms, and the sum is 0.

One nice consequence is that the number of terms in the sum is simply B − a rather than b − a + 1. Perhaps more importantly, we have this theorem:

Σ	+	Σ	=	Σ	,
A≤i<B		B≤i<C		A≤i<C

which looks nicer than

b		c		c
Σ	+	Σ	=	Σ	.
i=a		i=b+1		i=a

The formulas for summing cubic polynomials also tend to be slightly simpler. With the traditional numbering, we have these (from the textbook):

The sum, from i = 0 to i = b, of a constant c is c(b + 1).
The sum, from i = 0 to i = b, of i itself is b(b + 1)/2.
The sum, from i = 0 to i = b, of i² is b(b + 1)(2b + 1)/6.
The sum, from i = 0 to i = b, of i³ is b²(b + 1)²/4.

With the off-by-1 numbering, we have these:

The sum, over 0 ≤ i < B, of a constant c is cB.
The sum, over 0 ≤ i < B, of i itself is B(B − 1)/2.
The sum, over 0 ≤ i < B, of i² is B(B − 1)(2B − 1)/6.
The sum, over 0 ≤ i < B, of i³ is B²(B − 1)²/4.

The upshot of all of this is that, when you see (for example) a sum as i runs from 2 to 5, you might want to think of it as a sum over 2 ≤ i< 6 instead.

It's also handy to have more general formulas for summing cubic polynomials, starting at an arbitrary place rather than at i = 0. With the traditional numbering, we have these:

The sum, from i = a to i = b, of a constant c is c(b − a + 1).
The sum, from i = a to i = b, of i itself is (a + b)(b − a + 1)/2.
The sum, from i = a to i = b, of i² is (2a² + 2ab + 2b² − a + b)(b − a + 1)/6.
The sum, from i = a to i = b, of i³ is (a² + b² − a + b)(a + b)(b − a + 1)/4.

With the off-by-1 numbering, we have these:

The sum, over A ≤ i < B, of a constant c is c(B − A).
The sum, over A ≤ i < B, of i itself is (B − A)(A + B − 1)/2.
The sum, over A ≤ i < B, of i² is (B − A)(2A² + 2AB + 2B² − 3A − 3B + 1)/6.
The sum, over A ≤ i < B, of i³ is (B − A)(A + B − 1)(A² + B² − A − B)/4.

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