Introduction

Say we open a bank account on September 1st in which we deposit x dollars every day for the entire month, where x is equal to the calendar date. This means we deposit $5 on the 5th, $12 on the 12th etc. The total amount in the account on the n-th day is 1 + 2 + 3 + .... + n or the sum of the first n positive integers.

The symbolic representation of this sum is

If we want the sum of the squares of the first n integers we use

Sigma Notation and Summation Formulae & Theorems

The symbol is the Greek letter sigma. It acts as a summation operator. The teeny numbers above and below the sigma are called indices (the plural of index) and they define the extent of the summation -- where to start and end the summing. Pay close attention to the indices of summation. They don't always start at 1. Most books use i, j, or k to represent the index of summation.

This , read the summation of i from one to five -- equals the sum of the first 5 positive integers, or {1 + 2 + 3 + 4 + 5 = 15}.
We could use a calculator each time to find such sums, however when summing to 50, it would take too long. We use these formulas and theorems to find them.

Summation Formulae:

If n is a positive integer or counting number, then these formulae apply:

a)	b)	c)
I wrote the denominator of formula (b) as 2 times 3 so that we see formula (a) which we already know, multiplied by (2n + 1) / 3. It's just a memory aid.

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The expression on the left side of these formulae (with the sigma) is called the open form of the summation, the expression on the side right is the closed form.

Summation Theorems:

These theorems tell us how to work with the sums of constants etc.

Given 2 sets of real numbers {a₁ , a₂ , a₃ , ........ a_n} and {b₁ , b₂ , b₃ , ........ b_n},
a constant c, and n, a positive integer, these 5 theorems apply:

a)	b)
c)	d)
e) (notice the indices don't start at 1)

Examples:

1/ By formula (a) .

2/ By formula (b), .

3/ By formula (c), .

4/ Change to its closed form and find an algebraic expression for the sum.

Since (3 + i) ² = 9 + 6i + i² , we need to evaluate

Using the formulae, we get .

After some algebra to reduce and find the common denominator, we get:

5/ Given {a₁ , a₂ , a₃ } = { 2, 7, 12 } , .

6/ .

7/ .

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8/ .

9/

10/ In some cases, none of the formulae apply so we just find the sum of the terms.

If asked to find , we add the 4 terms: ( – 1) + (1) + (1/3) + (1/5) = 8/15.

Zero as an Index: If we start at 0 instead of 1 we must be sure to include the zero-th term.

When summing a constant such as , we will not get 4 × 7 since we need to add a 7 for the 0th, 1st, 2nd, 3rd and 4th terms -- which means 5 × 7.

So, we can state that .

In a case such as , since i² = 0 when i = 0 , only the summation of 2 will change. We can rewrite as .

Practice

1/ Use the summation theorems and formulae to evaluate:

a)

b)

c)

d)

2/ Express these sums in Sigma notation:

a) 1 + 2 + 3 + . . . . + 10	b) 3 $ 1 + 3 $ 2 + 3 $ 3 + . . . + 3 $ 20
c) 1 $ 2 + 2 $ 3 + 3 $ 4 + . . . + 49 $ 50	d) 1 + 3 + 5 + 7 . . . . + 15

3/ Express 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ in sigma notation with:

a) j = 0 as lower limit of summation

b) j = 2 as lower limit of summation

4/ Express in sigma notation with:

a) k = 0 as lower limit of summation

b) k = 5 as lower limit of summation

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Solutions

1/

a)	b)
c)	d)

2/

a) 1 + 2 + 3 + . . . . + 10 =	b) 3 $ 1 + 3 $ 2 + 3 $ 3 + . . . + 3 $ 20 =
c) 1 $ 2 + 2 $ 3 + 3 $ 4 + . . . + 49 $ 50 =
d) 1 + 3 + 5 + 7 . . . . + 15 =

3/ a) 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ in sigma notation with j = 0 is .

b) 1 + 2 + 2² + 2³ + 2⁴ + 2⁵ in sigma notation with j = 2 is .

4/ in sigma notation with:

a) k = 0 as lower limit of summation Since the lower limit is k -4 , so is the upper limit. To maintain equality, we must raise the summation variable expressions by 4. We get:

b) k = 5 as lower limit of summation. Using the same reasoning, we get: .

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