Li Li
208190233
12/12
Exercise 1: 2/2
(a)
Fibonacci Sequence:
f n  f n1  f n2 , f1  1, f 2  1, n  2, n  N ( Natural Numbers)
So, f1  1
f2  1
f 3  f 2  f1  1  1  2 ;
f4  f3  f2  2 1  3 ;
f5  f 4  f3  3  2  5 ;
f6  f5  f4  5  3  8 ;
f 7  f 6  f 5  8  5  13 ;
f 8  f 7  f 6  13  8  21 ;
(b)
fn
f
5
2
 5  1
f n1 f 4 3
3
f
f
8
3
If n=6, n  6   1
f n1 f 5 5
5
f
f
13
5
If n=7, n  7 
1
f n1 f 6
8
8
If n=5,
Notice that as n gets larger and larger this quotient when expressed as a
decimal gets closer and closer to the decimal representation of
1 5
 1.61803398875 . We see that
2
f8
f
233
 21 / 13  1.615384 , 13 
 1.6.8055
f7
f12 144
lim n
fn
fn 1

1 5
2
and it is true that
(b)
x2  x 1  0
 (1)  (1) 2  4  1  (1) 1  1  4 1
5
x

 
2 1
2
2 2
So, x has two roots:
1
5
2.24
 0.5  1.12  1.62 (as decimals)
x1  
(in terms of square root), or x1  0.5 
2
2 2
1
5
2.24
 0.5  1.12  0.62 (as
x2  
(in terms of square root), or x 2  0.5 
2
2 2
decimals)
Should have shown more decimal places
Exercise 2: 2/2
(a)
So,
an 
1
, a1  1
1  a n 1
1
1
1
1


 ;
1  a 21 1  a1 1  1 2
1
1
1
1 2
a3 


  ;
1 3 3
1  a31 1  a 2
1
2 2
1
1
1
1 3
a4 


  ;
2 5 5
1  a 41 1  a3
1
3 3
1
1
1
1 5
a5 


  ;
3 8 8
1  a51 1  a 4
1
5 5
1
1
1
1
8
a6 




5 13 13
1  a61 1  a5
1
8
8
a2 
1 5
then it follows that  2    1  0 since
2
2
 is a root of the equation x  x  1  0 , but by some algebraic
1
1
manipulation we see that   1  And our sequence above has limit
Note that if we set  


(b) a n 
2
3n
So, a n is a geometric sequence where a 0 
2
1
 2, r 
0
3
3

Let S4 be the 4th partial sum of the series
a
k 0
n
, and S4 denote the sum of the first 5 terms
1
1
1
2  2  ( ) 41 2(1  ( ) 5 ) 1  ( ) 5
3
3
3  3  ( 1 ) 4  3  1  2 80
Then, partial sum S 4 


1
2
1
3
81
81
1
3
3
3
Alternatively, we can S4 by adding the first 5 terms:
2
a0  0  2
3
2 2
a1  1 
3
3
2 2
a2  2 
9
3
2
2
a3  3 
27
3
2
2
a4  4 
81
3
Then, partial sum
2 2 2
2
2  27  2  9  2  3  2
54  18  6  2
80
S4  2   

 2
 2
2
3 9 27 81
81
81
81
(c) Note that this change in the problem reflects the
interchange in the Discussion Forum - all the more reason to
interact in this way
4
1
3
k 2
6
k
2

1
1
1 9  3  1 13
 3 4 

2
81
81
3
3
3
2
2
2
2
2
2
2
2
1
2
1
2
13
 n  1  0  1  1  1  2  1  3  1  4  1  5  1  6  1  2  1  3  2  5  3  7  5 70
n 0
Exercise 3: 2/2
(a)
Arithmetic sequence: 11, 8, 5, 2…
As 8-11=-3, 5-8= -3, 2-5=-3, the common differenced=-3
Then, the fifth term is: 11+(5-1)(-3)=11-12 = -1 (or we can get the fifth term from 4th
term: 2+(-3)=-1);
The nth term is:11+(n-1) (-3)=11+3-3n = 14-3n
The 100th term is:11+(100-1) (-3)=11-99  3 = 11 (1-27) = -26 11= -286
(b)
Arithmetic sequence: a =3, d =2, n =12
12(12  1)  2
 36  12  11  168
Then, Sn=S12=123+
2
(c)
Arithmetic sequence: a=50,000, d= 2800, n=15
15(15  1)  2800
 750,000  15  14  1400  1,044,000
So, S15=50,00015+
2
Exercise 4: 2/2
(a)
1. Geometric Sequence: 1, 2 , 2, 2 2 ,…
As the 2nd term divided by the first term is 2 , the 3rd term divided by the 2nd term
is 2 , and the 4th term divided by the 3rd them is also 2 ,
the common ratio is: 2
Then, the fifth term is: 1 ( 2 )  ( 2 )  4 ,( or we can get it from 4th term:
4
4
2 2  2 = 4)
The nth term is: 1 ( 2 ) n 1 = ( 2 ) n 1 ;
1
1
2. Geometric Sequence: -8, -2,  ,  ,…
8
2
1
As the 2nd term divided by the first term is , the 3rd term divided by the 2nd term is
4
1
1
, and the 4th term divided by the 3rd them is also
4
4
1
the common ratio is:
4
1
8
1
  , or we can get it from 4th term
Then, the fifth term is: -8  ( ) 4  
4
256
32
1 1
1
  =8 4
32
1
The nth term is: -8  ( ) n 1
4
(b)
2
1
,r  ,n  4
3
3
2
1
2
1
(1  ( ) 5 )
(1  ( ) 5 )
a  ar 41 3
1
1
242
3
3
Then, the sum S 4 

 3
 1  ( )5  1 

1
2
1 r
3
243 243
1
3
3
Geometric sequence: a 
Exercise 5: 2/2
1. Geometric series: 1-
1 1 1
+ –
+…
3 9 27
1
As the 2nd term divided by the first term is - , the 3rd term divided by the 2nd term is
3
1
1
1
- , and the 4th term divided by the 3rd them is also - , the common ratio is: 3
3
3
So, the sum of this infinite series is: S n  lim S n n 
2. Geometric series: 3-
1
1
1  ( )
3

1 3

4 4
3
3 3 3
+ – …
2 4 8
1
, the 3rd term divided by the 2nd term is
2
1
1
1
- , and the 4th term divided by the 3rd them is also - , the common ratio is: 2
2
2
As the 2nd term divided by the first term is -
So, the sum of this infinite series is: S n  lim S n n 
3
1
1  ( )
2

3
2
3
2
Exercise 6: 2/2 A good exposition
A ball is dropped from a height of 80feet, and the height of each rebounce is
2
of that of
3
the previous drop, so the ball will travel like this:
1st
dro
p
80
1st
rebounce
80 
2
3
2nd drop
80 
2
3
2nd
rebounce
3rd drop
3rd
rebounce
4th drop
4th
rebounce
5th drop
nth
rebounce
(n+1)th
drop
2
2
2
2
2
2
2
2
80  ( ) 2 80  ( ) 2 80  ( ) 3 80  ( ) 3 80  ( ) 4 80  ( ) 4 80  ( ) n 80  ( ) n
3
3
3
3
3
3
3
3
2
,
3
So, on the fifth bounce, the height of the ball is:
2
32 2560
130
80  ( ) 5  80 

 10
feet
3
243 243
243
It is clear that the ratio is
2
on the nth bounce, the height of the ball is: 80  ( ) n feet
3
The distance it travels up and down before it comes to a rest is:
2
2
2
2
2
2
2
2
80+ 80  + 80  + 80  ( ) 2 + 80  ( ) 2 + 80  ( ) 3 + 80  ( ) 3 + 80  ( ) 4 + 80  ( ) 4 +…
3
3
3
3
3
3
3
3
2
2
80  ( ) n + 80  ( ) n +…
3
3
2
2
2 2
2 2
2 3
2 3
2 4
2 4
=80+80+ 80  + 80  + 80  ( ) + 80  ( ) + 80  ( ) + 80  ( ) + 80  ( ) + 80  ( ) +
3
3
3
3
3
3
3
3
2 n
2 n
…+ 80  ( ) + 80  ( ) - 80+…
3
3
2
2
2
2
2
=2(80+ 80  + 80  ( ) 2 + 80  ( ) 3 + 80  ( ) 4 +… + 80  ( ) n +…) – 80
3
3
3
3
3
2
2 2
2 3
2 4
2 n
As 80+ 80  + 80  ( ) + 80  ( ) + 80  ( ) +… + 80  ( ) +… is a geometric series
3
3
3
3
3
2
where a=80, r= ,
3
2
2
2
2
2
the sum of 80+ 80  + 80  ( ) 2 + 80  ( ) 3 + 80  ( ) 4 +… + 80  ( ) n +…
3
3
3
3
3
80
80

 240
is S n  lim S n n 
2
1
1
3
3
Then, the distance it travels up and down before it comes to a rest is:
2  240 – 80 = 400 feet