Difference method for some finite algebraic series
Yue Kwok Choy
(A) If the general term, ur , of the series is in "product" form, you can add one more factor to the end
of the general term ur , so as to form a sequence vr and then apply the difference method.
Example 1
Find the sum of n terms of the series :
1  4  7  4  7  10  7  10  13  ...
Solution
(1) Find the general term of the given series:
,r1.
u r  3r  23r  13r  4
Note : The general term is not u r  rr  3r  6
(2) Form another sequence vr by adding one more factor to the end of the general term ur :
,r1.
v r  3r  23r  13r  43r  7
(3) Find
vr-1 :
v r1  3r  53r  23r  13r  4
,r2.
(4) Form the difference :
vr  vr1  3r  23r  13r  43r  7  3r  53r  23r  13r  4
 3r  23r  13r  43r  7   3r  5
 123r  23r  13r  4  12u r
,r2.
(5) Find the sum :
12 u r  12  u r  12u1   v r  v r 1   12u1 , note that the difference is good for r  2 .
n
n
n
r 1
r 2
r 2
 v n  v n 1   v n 1  v n 2   ...v 3  v 2   v 2  v1   12u1
 v n  v1   12u1
 3n  23n  13n  43n  7  1  4  7  10  121  4  7
n
 ur 

r 1
1
3n  23n  13n  43n  7   56
12
(6) It is more convenient to define v0 . Since v1 – v0 = 12 u1 , we have:
1  4  7  10  v 0  12  1  4  7 and we define v 0  56
(or  2  1  4  7) .
12 u r   v r  v r 1   v n  v 0  3n  23n  13n  43n  7   56

n
n
r 1
r 1
n
 ur 
Exercise :
r 1
1
3n  23n  13n  43n  7   56
12
Show that :
1  3  5  3  5  7  5  7  9  ...(to n terms)  nn  22n 2  4n  1
1
(B) If the general term, ur , is in "quotient" form, you can remove one more factor at the end of the
general term ur , so as to form a sequence vr and then apply the difference method.
Example 2
Find the sum of n terms of the series :
Solution
(1) Find the general term of the given series:
1
ur 
3r  23r  13r  4
1
1
1


 ...
1  4  7 4  7  10 7  10  13
,r1.
(2) Form another sequence vr by removing one factor at the end of the general term ur :
1
,r1.
vr 
3r  13r  4
(3) Find
v r 1 
vr-1 :
Note that
1
4
v0 
denominator.
1
3r  23r  1
,r1
is well-defined and that is why in
(2), we remove the first factor in the
Removing (3r + 4) instead in the denominator will get
v0
undefined .
(4) Form the difference :
v r  v r 1 
1
1

3r  13r  4 3r  23r  1

1
3r  2  3r  4
3r  23r  13r  4
 6u r
,r2.
(5) Find the sum :
 6 u r   v r  v r 1   v n  v 0 

n
n
r 1
r 1
1
1

3n  13n  4 4
n

1 1
1
 ur   
r 1
6  4 3n  13n  4
Exercise :
Prove that :
1
1
1
1
1

 ... 
 
(1)
1 2  3 2  3  4
n n  1n  2 4 2n  1n  2
n
(2) 
r 1

1
1 1
1
  
2r  12r  12r  3 4  3 2n  12n  3
2
(C) More complicate finite series can be broken into two (or more) series and each can be handled
by the methods above .
Example 3
Find the sum of n terms of the series :
1 4  7  8  4  7 10 11  7 10 13 14  ...
Hint
The general term,
u r  3r  23r  13r  43r  5
 3r  23r  13r  43r  7   2
 3r  23r  13r  43r  7  23r  23r  13r  4
1
3n  23n  13n  43n  7 2n  5  28
10
Answer
Example 4
Find the sum of n terms of the series :
3
5
7


 ...
2 3 4 5 3 4 5 6 4 5 6  7
Hint
The general term,
ur 
2r  1
r  1r  2r  3r  4

Answer
2r  1  1
2
1


r  1r  2r  3r  4 r  2r  3r  4 r  3r  4
5 1
3n  5

72 3 n  2n  3n  4
(D) If the general term, ur , of the series is in "product + quotient" form with same number of factors
in numerator and denominator, you can add one more factor to the end of the general term ur ,
so as to form a sequence vr and then apply the difference method.
Example 5
Find the sum of n terms of the series :
2 24 246


 ...
1 1 3 1 3  5
2  4  ...  2r 
2  4  ...  2r 2r  2
, vr 
1  3  ...  2r  1
1  3  ...  2r  1
Hint
ur 
Answer
1  2  4  ...  2r 2n  2 
 2

3  1  3  ...  2n  1

3