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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

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appendix E Sigma Notation A35

2 Theorem If c is any constant (that is, it does not depend on i), then

(a) o n

ca i − c o n

a i

i−m

i−m

(c) o n

sa i 2 b i d − o n

a i 2 o n

b i

i−m

i−m i−m

(b) o n

sa i 1 b i d − o n

a i 1 o n

b i

i−m

i−m i−m

Proof To see why these rules are true, all we have to do is write both sides in

expanded form. Rule (a) is just the distributive property of real numbers:

ca m 1 ca m11 1 ∙ ∙ ∙ 1 ca n − csa m 1 a m11 1 ∙ ∙ ∙ 1 a n d

Rule (b) follows from the associative and commutative properties:

Rule (c) is proved similarly.

sa m 1 b m d 1 sa m11 1 b m11 d 1 ∙ ∙ ∙ 1 sa n 1 b n d

− sa m 1 a m11 1 ∙ ∙ ∙ 1 a n d 1 sb m 1 b m11 1 ∙ ∙ ∙ 1 b n d

EXAMPLE 3 Find o n

1.

SOLUTION

i−1

o n

1 − 1 1 1 1 ∙ ∙ ∙ 1 1 − n ■

i−1

n terms

EXAMPLE 4 Prove the formula for the sum of the first n positive integers:

o n

nsn 1 1d

i − 1 1 2 1 3 1 ∙ ∙ ∙ 1 n −

i−1

2

SOLUTIOn This formula can be proved by mathematical induction (see page 72) or

by the following method used by the German mathematician Karl Friedrich Gauss

(1777–1855) when he was ten years old.

Write the sum S twice, once in the usual order and once in reverse order:

S − 1 1 2 1 3 1 ∙ ∙ ∙ 1 sn 2 1d 1 n

S − n 1 sn 2 1d 1 sn 2 2d 1 ∙ ∙ ∙ 1 2 1 1

Adding all columns vertically, we get

2S − sn 1 1d 1 sn 1 1d 1 sn 1 1d 1 ∙ ∙ ∙ 1 sn 1 1d 1 sn 1 1d

On the right side there are n terms, each of which is n 1 1, so

2S − nsn 1 1d or S −

nsn 1 1d

2

EXAMPLE 5 Prove the formula for the sum of the squares of the first n positive

integers:

o n

nsn 1 1ds2n 1 1d

i 2 − 1 2 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 n 2 −

i−1

6

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