1.

2 Summation notation
It is understood that the area of a geometric figure is the measure that, in some way, provides the
size of the region closed by the figure. The area of a region in a plane is defined if the region is bounded
through a curve. In the study of the area, sums of many terms will be addressed, so an introduction is made to a
notation, called Sigma Notation or Summation Notation, to facilitate the writing of these sums. This
notation requires the use of the symbol∑, the uppercase sigma letter of the Greek alphabet.

Examples:

∑ i 2 = (1+)2(2+)2( 3)(
2 + )(
24+)25= 1 + 4 + 9 + 16 +25

=1

8
1 1 1 1 1 1 1
∑ = + + + + +
k 3456 78
=3

∑ ( 3i +2= ) 31 [ +(2+
) 32+] 2=[ 5+
( 8=
) 13 ]
=1

Definition of sigma notation

∑ () = ( () +) ( + 1+) +( 2+ ⋯)+ n - 1 +( )
=

Where y they are integers, and ≤ .

The number is called the lower limit of the sum, and is called the upper limit of the sum. The symbol
It receives the name of the index of the sum. This is a "fictitious" symbol because any other letter can be used.
for this purpose. For example,

Solver

6
2 ( 3)2 ( 4)2 ( 5)2 ( 6)2 9 162536
∑ = + + + = + + +
i + 1 3 + 14 + 1 5+ 1 6 + 1 4 5 6 7
=3
The following theorems deal with the use of sigma notation, and are useful for certain calculations. Where is
any constant.

∑ ∙ () = ∑ ()
=1 =1

∑ [ ( ) + ( )] =∑ ( ) +∑ ()
=1 1 =1

∑ = ∙
=1

( + 1) 2+ 1 1
∑ = = = 2+
2 2 2 2
=1

2
( + 1)(2n + 1 ) 2 3+ 3n2+ 1 1 2 1
∑ = = = 3+ +
6 6 3 2 6
=1

2(
3
+ 1) 2 4+ 2 3+ 2 1 4+
1 3 12
∑ = = = +
4 4 4 2 4
=1

( 2
4
+ 1)(2n + 1 3n)( +3n -1 ) 6n5+ 15 4+ 10n3− 1 5+
1 4 13 1
∑ = = = + −
30 30 5 2 3 30
=1
Example 1. Calculate the sum.

∑ ( 3i -2 )
=1

It is used

∑ ( 3i -2= ) ∑ 3i - ∑ 2
∑ [ () + ( )] = ∑ () +∑ ()
=1 =1 =1
=1 =1 =1

It is used

=3 ∑ −∑ 2 ∑ ∙ () = ∑ ()
=1 =1
=1 =1

It is used It is used
1 2 1 1 1
= 3( + )−2 ( ) ∑ 2
∑
2 2 = + = ∙
2 2
=1 =1

3 2+
3
= −2
2 2

3 2−
1
=
2 2

∑ ( − )= −
=
Example 2. Calculate the sum.

∑ [( i - 14i)(+ 3 )]
=1

Simplify
∑ [( i - 14i)(+ 3 = )] ∑ ( 4i 2i-- 3 )
) 3i - 4i2 - 3 = 4i -i - 3
( i - 14i)(+ 3 = 4i+ 2
=1 =1

It is used

= ∑ 4i 2 - ∑ −∑ 3 ∑ [ () + ( )] = ∑ () +∑ ()
=1 =1 =1
=1 =1 =1

It is used
2
=4 ∑ −∑ −∑ 3 ∑ ∙ () = ∑ ()
=1 =1 =1
=1 =1

It is used It is used It is used

1 3 1 2
1 1 1 13 12 1 1 1
= 4( + + )−( + ) -23 ( ) ∑ 2= + + ∑ = 2+ ∑ = ∙
3 2 6 2 2 3 2 6 2 2
=1 =1 =1

4 3+ 2+
2 12 1
= 2 − − - 3n
3 3 2 2

4 3+
3 2 17
= −
3 2 6

∑ [( − )( + )]= + −
=
Example 3. Calculate the sum.

∑ ( 2i - 3 ) 2
=1

Simplify
∑ ( 2i - 3 =) 2 ∑ ( 4i 212i
- +9 )
( 2i - 3 =) 24i2- 12i + 9
=1 =1

It is used

= ∑ 4i 2 - ∑ 12 +∑ 9 ∑ [ () + ( )] = ∑ () +∑ ()
1 =1 =1
=1 =1 =1

It is used
2
=4 ∑ − 12∑ +∑ 9 ∑ ∙ () = ∑ ()
=1 =1 =1
=1 =1

It is used It is used It is used

1 3
1 2
1 1 1 13 12 1 1 1
4 (+ + − 12 ) + +( 9 n2 ) ( ) ∑ 2= + + ∑ = 2+ ∑ = ∙
3 2 6 2 2 3 2 6 2 2
=1 =1 =1

4 3+ 2+
2
= 2 -6n2-6n + 9n
3 3

4 3-4n2+
11
=
3 3

∑ ( − )= − +
=