Solutions

Problem Set #7:

 

In each case provide one expression for the series given. Do not oversimplify.

 

 

1.Please sum the following expressions:

 

a)      3 + 6 + 9 + 12 + 15 + . . . + 300  = 3(1+2+3+4+ . . 100) = 3(100)(101)/2   (arithmetic)

b)     1 + 1/3 + 1/9 + 1/27 + . . . + 1/2187   = 1+ (1/3) + (1/3)2 + . . . + (1/3)7  = (1 – (1/3)8)/(1 – 1/3)  (geometric)

c)       5 + 6 + 7 + 8 + … + 80   =  -(1+2+3+4)  +  (1+2+3+4+5 + 6 + 7 + 8 + … + 80   ) = -10 +  (80)(81)/2 = -10 +3240 = 3230

d)     1/5 + 1/25 + 1/125 + 1/625 + 1/3125 + 1/15625  =  - 1  +  (1 + 1/5 + 1/25 + 1/125 + 1/625 + 1/3125 + 1/15625  )

= -1  +  (1- (1/5)7)/(1 – 1/5)   (geometric)

 

e)      3 +  3/2 + 3/4 + 3/8 + 3/16 + 3/32 + 3/64 + 3/128 + 3/256 + 3/512 + 3/1024 = 3( 1 + ½ + . . . + (1/2)10) = 3( 1 – (1/2)11)/(1-1/2)

 

2. Please  sum the following series:

 

a)  1 + 1/3 + 1/9 + . . . + 1/729    = (1 – (1/3)8/(1-  1/3)    r=1/3

 

b)   1 – 1/2 + 1/4 -1/8 + 1/16 . . . + 1/1024   = (1 – (-1/2)11 )/(1 – -½)    r = -1/2

 

c)  5 + 5/2 + 5/4 + . . . + 5/256   = 5( 1 + ½ + (1/2)2 + . . . + (1/2)8)  = 5( 1 – (1/2)9)/(1-  ½)

 

d)  1/25 + 1/125 + 1/625 + . . . + 1/78125  =  -1 -1/5 + (1 + 1/5 + 1/25 + 1/125 + 1/625 + . . . + 1/78125  ) = -6/5 +  geometric for r =1/5, first 7 terms

                                                                        = -6/5  +  (1 – (1/5)8) /(1- 1/5)

 

e)  6 + 8 + 10 + . . . + 100  = 2(3+4 + 5 + . . . + 50)  =  2[  -1 -2 + (1 + 2 + 3 + 4 + 5 + . . . +50) ]  = 2[ -3  +  (50)(51)/2 ] = 2[-3 + 25*51]

                                                                                                                                                            = 2[ -3 + 1275] = 2(1272) = 2544

 

3. Consider the following series:

                

 

  1. write out the first three terms of the series and also the last two = (1 – 4) + (4 -9) + (9 – 16) + . . . + (992 – 1002) + (1002 – 1012)

      =  1 - 1012   (note many cancellations)

  1. see if you can sum the series

 

4.  Let's look at the odd positive integers and add them up

 

      1 + 3 + 5 + 7 + . . . +33        for example

 

      a) can you find a method for adding up the first 17 odd numbers? (other than simply adding them one by one)

 

b)      can you extend your method and find a formula for the sum of the first  n odd integers?

 

1+3+5 = 9       1+3+5+7=16     1+3+5+7+9=25    1+3+5+7+9+11 = 36  ….. pattern seems clear.  The answer keeps coming up as the square of an integer.  Specifically, if you look at the patters, the sum of the first n odd integers is n2.

 

So apparently the sum of the first 17 odd integers would be 172   which is 289.

 

 

 

5.  A very hard ball is dropped from a height of 100 feet. It strikes the ground and bounces back up, each time reaching a height of 3/4 the height of the previous bounce. It bounces 12 times before coming to rest. What is the total distance it traveled? What does this have to do with series??

 

      drop    bounce 1           bounce 2                   bounce 3               bounce 4

 =  100 + (75  + 75)  +  ( (3/4)75 + (3/4)75) + ( (3/4) ( (3/4)75 + (3/4)75) + (3/4) ( (3/4)75 + (3/4)75) ) + . . .    

 

 

=   100 + (2)(3/4)100  +  2(3/4)(3/4)100 + 2(3/4)(3/4)(3/4)100 + . . .                   

 

=   100 [ 1  +  2(3/4)  + 2(3/4)2  + 2(3/4)3 + . . .          +  2(3/4)12]               note the power of ¾  goes with the  bounce number

 

so it appears we need to sum the first  12 powers of ¾   and then double that sum add 1 and multiply the whole thing by 100

 

 

To do this:  [2(3/4)  + 2(3/4)2  + 2(3/4)3 + . . .            2(3/4)12]  = 2[(3/4)  + (3/4)2  + (3/4)3 + . . .            (3/4)12]

 

                                                                                                    = 2(3/4) [ 1 + (3/4) + (3/4)2 +  . . . (3/4)11]

 

                                                                                                    = 2(3/4) ( 1 –(3/4)12)/(1 – (3/4) )  using the Geometric Series Formula

 

                                                                                                     =  (3/2)(4) (1 – (3/4)12)  =  6(1 – (3/4)12)

 

So the solution to the problem is

 

                        100(  1  +  6(1 – (3/4)12)  )  =  100 + 600( 1- (3/4)12 )  = 680.9941