4 Jan 1999

### he Notation

In mathematics, we often use symbols to simplify mathmatical expression.

Expression like 12 + 22 + 32 + ¼ + n2 is often written in short as år2, which means "the sum of all terms like r2 ".

To be more precise, numbers are placed above and below å to show where the series begins and ends.

Thus

 12 + 22 + 32 + ¼ + n2 = n å r = 1 r2.

In general, a finite series am + am + 1 + ¼ + an, where m £ n, is written as

 n å r = m ar .

In this notation,

• r is called the index of summation (may be replaced by other symbols),
• ar is the general term of the series,
• m is the lower limit,
• n is the upper limit,
• å is a greek capital letter (read as sigma) which corresponds to S (for sum).

For infinite series am + am + 1 + ¼, we replace the upper limit by ¥.

So the representation is

 ¥ å r = m ar .

Usually, the lower limit is 1, but it should not be assumed to be always the case.

 n å r = 1 ar = a1 + a2 + ¼ + an .

### ome Results

 n å ar r = m
=
 n å ar r = 1
-
 m - 1 å ar r = 1
where n > m ³ 1

 n å (aUr + bVr) r = m
= a
 n å Ur r = m
+ b
 n å Vr r = m
where a, b are constants

 n å k r = 1
= nk    where k is a constant

 n å r r = 1
= ½n(n + 1)

 n å r2 r = 1
= n(n + 1)(2n + 1)/6

 n å r3 r = 1
= [n(n + 1)/2]2

 n å kr r = 1
=
 k(kn - 1) ¾¾¾¾ k - 1
where k ¹ 1

### xamples

Example 1:  Find

 n å (r + 1)(r + 2). r = 1

Solution:

 n å (r + 1)(r + 2) r = 1
=
 n å (r2 + 3r +2) r = 1

=
 n å r2 + 3 r = 1
 n å r + r = 1
 n å 2 r = 1

=  (n/6)(n + 1)(2n + 1) + 3(n/2)(n + 1) + 2n

=  (n/3)[n2 + 6n + 11].

Example 2:  Find

 2n å (2r - 3r2). r = n + 1

Solution:

 2n å (2r - 3r2) r = n + 1
=
(2n + 1  + 2n + 2  + ¼ + 22n) - 3[
 2n å r2 - r = 1
 n å r2 r = 1
]

=
 2n + 1(2n - 1) ¾¾¾¾¾¾ - 3[(2n/6)(2n + 1)(4n + 1) - (n/6)(n + 1)(2n + 1)] 2 - 1

=  22n + 1 - 2n + 1 - (n/2)(2n + 1)(7n + 1).

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