4 Jan 1999
he Notation
In mathematics, we often use symbols to simplify mathmatical expression.
Expression like 1^{2} + 2^{2} + 3^{2} + ¼ + n^{2} is often written in short as år^{2}, which means "the sum of all terms like r^{2} ".
To be more precise, numbers are placed above and below å to show where the series begins and ends.
Thus
1^{2} + 2^{2} + 3^{2} + ¼ + n^{2} = 
n
å
r = 1

r^{2}. 
In general, a finite series a_{m} + a_{m}_{ + 1} + ¼
+ a_{n}, where m £ n, is written as
In this notation,

r is called the index of summation (may be replaced by other symbols),

a_{r} 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 a_{m} + a_{m}_{ +
1} + ¼, we replace the upper limit
by ¥.
So the representation is
Usually, the lower limit is 1, but it should not be assumed to be always
the case.
n
å
r = 1

a_{r} 
= a_{1} + a_{2} + ¼ + a_{n} . 
ome Results



n


å

(aU_{r} + bV_{r}) 
r = m 


= a 

+ b 

where a, b are constants 



= nk where k is a constant 


= ½n(n + 1) 


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


= [n(n + 1)/2]^{2} 



xamples
Example 1: Find 
n


å

(r + 1)(r + 2). 
r = 1 


Solution:

= 
n


å

(r^{2} + 3r +2) 
r = 1 




= 



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


= 
(n/3)[n^{2} + 6n
+ 11]. 
Example 2: Find 
2n


å

(2^{r}  3r^{2}). 
r = n + 1 


Solution:
2n


å

(2^{r}  3r^{2}) 
r = n + 1 


= 
(2^{n + 1} + 2^{n + 2} + ¼ + 2^{2n})  3[ 


] 



= 
2^{n + 1}(2^{n}  1)


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





= 
2^{2n + 1}  2^{n
+ 1}  (n/2)(2n + 1)(7n
+ 1). 