Basic Sigma Notation and Rules

n

r 1 2 3 4 5 nr 1

(u ) u u u u u ......u

Addition rule:

where vr denotes rth term of sequence

Constant Multiple rule:

So

n

r 1

2r 1 2 1 1 2 2 1 2 3 1 ...... 2 n 1

So r starts at 1 and increases up to a finishing value of n

n n n

r r r rr 1 r 1 r 1

(u v ) u v

n n

r rr 1 r 1

ku k u

52 2 2 2 2 2

r 1

(r ) 1 2 3 4 5

5

2 2 2 2 2 2

r 1

(2r 1) 1 3 5 7 9

5

r 1

(2) 2 2 2 2 2 5 2

Using the addition and multiplication rules

5 52 2

r 1 1

5 5 52

1 1 1

5 5 52

1 1 1

(2r 1) 4r 4r 1

4r 4r 1

4 r 4 r 1

5

r 2 2 2 2 2 2

r 1

1 (r ) 1 2 3 4 5

5

r 1 2 2 2 2 2 2

r 1

1 (r ) 1 2 3 4 5

Changing the signs in a series

The term (-1)r means that all the odd terms in the series are –ve

The term (-1)r+1 means that all the odd terms in the series are +ve

n

r 1

k k n

3

r 1

5 5 3 15

n

r 1

1r n(n 1)

2

3

r 1

1r (3)(3 1) 6

2

n2

r 1

1r n(n 1)(2n 1)

6

3

2

r 1

1r (3)(3 1)(6 1) 14

6

n

3 2 2

r 1

1r n (n 1)

4

3

3 2 2

r 1

1r (3) (3 1) 36

4

i.e 1 + 2 + 3 = 6

i.e 12 +22 + 32 = 14

i.e 13 + 23 33 = 36

i.e 5 + 5 + 5 = 15

Difference Method1) Express the expression g(r) you are trying to sum as

2) Substitute values from 1 to n into this expression and determine the sum after cancelling the relevant terms.

g(r) = f(r+1) – f(r) or f(r) – f(r+1)

1 1 1

r(r 1) r r 1

This is in the form g(r) = f(r) - f(r+1)

f(r) = 1

r 1

r 1f(r+1) =

1 1 1 (r 1) 1 r

r r 1 r(r 1)

r 1 r

r(r 1)

1 =

r(r 1)

1g(r)=

r(r+1)

So g(r) = f(r) – f(r+1)

with g(r) = , f(r) = and f(r + 1) =

1 1 1

r(r 1) r r 1

1

r 1

r 1

1

r(r+1)

1 1 1

so r(r 1) r r 1

Using successive r values make a table of results

r = 1

1 1

1 2

r = 2

1 1

2 3

r = 3

1 1

3 4

r = 4

1 1

4 5

r = n 1 1

n n 1

1 1 1

so r(r 1) r r 1

1

n 1n

n 1The terms which remain are 1 - which simplifies to

Questions involving FactorialsExam qu. Show that (r + 2)! – (r + 1)! = (r + 1)2 r!

Hence find 221! + 32 2! + 42 3! …………(n + 1)2 n!

(r + 2)! = (r + 2)(r + 1)(r )(r – 1)…….. = (r + 2)(r + 1)r!

= r!(r+1)2

(r + 1)! = (r + 1) (r )(r – 1)…….. = (r + 1)r!

So (r + 2)! – (r + 1)! = (r + 2)(r + 1)r! – (r + 1)r!

= (r)!(r+1)((r + 2) –1)

n

r

n

r

)!r( )!r(!r)r( !n)n(... ! !11

2222 12112312

Using successive r values make a table of results

r=1 3! –2! r=2 4! –3! r=3 5! –4! r=4 6! –5! r=n–1 (n+1)! –n! r=n (n+2)! –(n+1)!

n

1r

n

1r

2 )!1r( )!2r(!r)1r( = (n + 2)! – 2!