n 1
a1  21  5  7
n  2 a2  22  5  9
n  3 a3  23  5  11
n  4 a4  24  5  13
n5
a5  25  5  15
7,9,11,13,15
n 1
n2
n3
n4
n5
1

 1
1
a1 

21  1 3
2

 1
1
a2 

22  1 5
3

 1
1
a3 

23  1 7
4

 1
1
a4 

24  1 9
5

 1
1
a5 

25  1 11
1 1 1 1 1
, , , ,
3 5 7 9 11
Alternating signs
means we have
(-1)n-1 power.
Make a T-chart and find a pattern
with the difference tests.
n
an
1
1
2
3
3
5
4
7
an  2 n  1
n
an
2 1  1
4 1  3
1
2
2
5
6 1 5
8 1 7
3
10
2
2
2
1st level
Difference Test.
Linear equation
with a slope of 2.
What needs to be
done to 2, 4, 6, 8
to get 1, 3, 5, 7?
an  2 n  1
4
17
an  n 2  1
1 1  2
2 4 1 5
2 9  1  10
3
5
7
16  1  17
2nd level Difference What needs to be
done to 1, 4, 9, 16
Test. Quadratic
to get 2, 5, 10, 17?
equation
an   1
n 1
n
2

1

a0  1
1, 1, 2, 3, 5, 8, 13, 21,...
a1  1
 
n  2 a2  a22  a21  a0  a1  1  1  2
n  3 a3  a32  a31  a1  a2  1  2  3
n  4 a4  a42  a41  a2  a3  2  3  5
Do you see a pattern?
0! 1
1! 1
2! 2 1  2
3! 3  2 1  6
4! 4  3  2 1  24
5! 5  4  3  2 1  120
6! 6  5  4  3  2 1  720
20 1
a0 
 1
0! 1
21 2
a1 
 2
1! 1
22 4
a2 
 2
2! 2
23 8 4
a3   
3! 6 3
2 4 16 2
a4  

4! 24 3
4
8! 8  7  6!

2!6! 2 1  6!
47

 28
1
2
2!6! 2!6  5!

3!5! 3  2!5!
2
 2
1
n!
n  n  1!

n  1! n  1!
n
 n
1
 31  32  33  34  35
 3  6  9 12 15  45

 
 
4
n
k 1
2
3
9
k
k 1
 
 1  32  1  42  1  52  1  62
 10 17  26  37  90
n
2
1 1 2 3
2
0 2 2
2
2
k 1

1
k 1
The k values are 1 bigger than the powers on the 2.
Notice that all the
properties creates
summations that all start at
k = 1 and ends with n.
Find each sum.
Property #2 and #6.
n=
 1010  1 
  3  511  165
k  3
3k  3 
2


k 1
k 1
10
10


Property #3.
8
k 1
Property #8.
4
Property #1.
 88  1 
 88  1 
k  1   k  1  
 1  
 18


 2  k 1
 2 
k 1
k 1

n=
5
3

8
8
2
8
3
2
2
 49   8  36  8  1296  8  1304
2
Property #2, #3, and #4.
n=
 k
24
k 1
 7k  2   k  7 k   2
24
2
24
24
k 1
k 1
2
k 1
4
Property #7, #6, and #1.
12
 2424  1224   1   2424  1 

  7
  224
6
2

 

 42549  71225  224  2848
Property #2.
 4k 
n = 20
k 6
2
Property #5.
 20 2 5 2 
 4  k  4 k   k 
k 6
k 1
 k 1

Not k = 1, Property #5
20
2
 2020  1220  1   55  125  1 
 4  


6
6
 


10 7
 202141   5611 
 4  


6
  6 

 4  10741  511  11260
1st level difference test.
n = 40
a1 = 2
+4 +4 +4 +4
d=4
a40 = 2 + 4(40 – 1)
a40 = 2 + 4(39) = 158
Treat the terms like points, as we did in Sect. 9.1, ( 4, 20 ) and ( 13, 65 ).
The difference, d, is the same as the slope. Find the slope between the 2 points.
65  20 45
d

 5 Find a1, use ( 4, 20 ) and ( 1, a1 ) with the slope formula.
13  4
9
20  a1
20  a1
20  a1  15 a1  5
5
5
4 1
3
an  5  5n  1
This is a visual approach to solving this problem. Make consecutive
blanks for each term.
d d d d d d d d d
a1
5 ____,
10 ____,
15 ____,
20 ____, ____, ____, ____, ____, ____, ____, ____, ____
65
____,
5
5
5
Make an equation from 20 to 65 with the 9 d’s.
Now subtract the d
value backwards
for the 1st term.
an  5  5n  1
20  9d  65
9d  45
d 5
n
a1
an
101
a1  an 
n
an  a1  d n  1
a1
d = -50
Solve for n.
Build the formula. Where did the
numbers come from?
n
These two formulas


S n  a1  an go hand in hand.
2
n
S n  a1  an 
+11 +11 +11
2
150
5  1644  751649  123,675
S150 
2
a1
n
S n  a1  an 
2
101 times how many groups?
101 ( 50 ) = 5050
n
2
+11
a150 = ?
d = 11
a150  5  11150  1
a150  1644
an
n=?
an  a1  d n  1  250  1000  50n 1
 750  50n 1 16  n
Find these values.
15  n 1
16
1000  250
2
16
 1250   81250 
2
S16  10,000
S16 
a1
n = 10
*3 *3 *3
*r
*r
r=3
an  a1r n 1
a10  2  3101  2  39  39,366
36r 2  16
16 4
2
r 

2
3
36 9
 or 
With
two
r
values
gives
us
2
equations.
3
2
2
r
 n 1
 n 1
 2
3
2
a

81



an  81 
n
 3
3
Make an equation from
81 ____,
54 ____,
36 ____, ____
16 36 to 16 with the 2 r’s.
____,
2
3
 or 
3
2
Multiply both sides by a –r.
This creates opposite terms.
S n  a1  a1r  a1r 2  a1r 3  ...  a1r n 2   a1r n 1
 rS n  a1r  a1r 2  a1r 3  ...  a1r n 2   a1r n 1  a1r n 
Add the equations together
and cancel terms.
Factor out the S.
S n  rS n  a1  a1r n

a1 1  r n
Sn 
1  r 
Divide by (1 – r).
n = 10

a1 1  r n
Sn 
1  r 

a1 = 5


S n 1  r   a1 1  r
r =2


n

Factor out the a1
5 1  210  51  1024   5 1023  5115
S10 
 1
 1
1  2
? n  ? a1  1000
r
)oo
(½
will continue to get
smaller until it is zero!
  1  
10001    
 2 


S 
1

1  
2


1
2
10001  0 
 10002  2000
 1  Dividing by ( ½ ) is the same as
 
 2  multiplying by the reciprocal of 2.

0.9  0.9  0.09  0.009  0.0009  ...
9
9
9
9




 ...
10 100 1000 10000
n
1
9
a1 
r
10
10
 1 
  0
 10 
S 
9   1 
1  

10   10 
1 

1



 10 


9


  10  1
9
10
0.9  1
Determine when a series diverge or converge.
If a finite sum S n approaches a number L as n   , we say the

k 1
a

r
infinite geometric series  1
converges. A series
k 1
diverges when the sum is  .
An infinite geometric series will always converge when 0  r  1 .
This gives us two formulas for geometric series.
a1
a1 1  r n
S 
Sn 
1  r 
1  r 


An infinite geometric series will always diverge when r  1.
2nd STAT – OPS #5 is the sequence formula.
seq(function,variable,starting value,ending value)
Find the first 5 terms of an  3n  5n  4
2
2nd STAT – MATH #5 is the sum formula.
Used for Series questions.
sum(seq(function,variable,starting value,ending value))
5
Find
2
3
n
  5n  4
n 1