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					                                   Geometric
                                 Sequences & Series
                                       By: Jeffrey Bivin
                                    Lake Zurich High School

                                      jeff.bivin@lz95.org




Last Updated: October 11, 2005
                     Geometric Sequences
    1, 2, 4, 8, 16, 32, … 2n-1, …

    3, 9, 27, 81, 243, … 3n, . . .

                                            n 1
    81, 54, 36, 24, 16,        … 81   2 , . .     .
                                        3
                               34 2 n 1   2 n 1
                                   n 1
                                             n 5
                                 3           3
Jeff Bivin -- LZHS
                            nth term
                     of geometric sequence


                       an =   a1·r (n-1)




Jeff Bivin -- LZHS
                     Find the nth term of the
                      geometric sequence
                            First term is 2
                          Common ratio is 3

                            an =   a1·r(n-1)


                            an =   2(3)(n-1)



Jeff Bivin -- LZHS
                      Find the nth term of a
                      geometric sequence
                            First term is 128
                          Common ratio is (1/2)
                     an = a1·r (n-1)
                                   n 1
                             1                      1
                     an  128 
                             2
                                          an         n 8
                                                  2
                             1 
                     an  2  n 1 
                           7

                            2 
Jeff Bivin -- LZHS
                     Find the nth term of the
                      geometric sequence
                            First term is 64
                          Common ratio is (3/2)
                     an = a1·r(n-1)
                                   n 1
                             3                  n 1
                      an  64                3
                             2          an  n  7
                                               2
                            3 
                               n 1
                     an  2  n1 
                           6
                            2 
                                 
Jeff Bivin -- LZHS
                     Finding the 10th term
                          3, 6, 12, 24, 48, . . .
    a1 = 3
    r=2                 an =  a1·r(n-1)
    n = 10                          10-1
                        an =  3·(2)
                        an = 3·(2)9

                        an = 3·(512)
                        an = 1536
Jeff Bivin -- LZHS
                     Finding the 8th term
                       2, -10, 50, -250, 1250, . . .
     a1 = 2
     r = -5            an =   a1·r(n-1)
     n=8                             8-1
                       an =   2·(-5)
                       an = 2·(-5)7

                       an = 2·(-78125)
                       an = -156250
Jeff Bivin -- LZHS
                             Sum it up

                             n
                                                a1  a1r   n
                     Sn    a r
                            i 1
                                   1
                                       n 1
                                              
                                                  1 r




Jeff Bivin -- LZHS
                 1 + 3 + 9 + 27 + 81 + 243
                                 a1  a1r
                                        n
                           Sn 
                                   1 r
            a1 = 1
                                1  1 3 6
            r=3            Sn 
            n=6                    1 3
                                1  729
                           Sn 
                                   2
                                 728
                           Sn          364
Jeff Bivin -- LZHS
                                  2
     4 - 8 + 16 - 32 + 64 – 128 + 256
                               a1  a1r n
                         Sn 
                                  1 r
            a1 = 4
                            4  4  ( 2 ) 7
            r = -2     Sn 
            n=7                1  ( 2 )
                            4  4(128)
                       Sn 
                                  3
                          4  512 516
                     Sn                     172
Jeff Bivin -- LZHS
                             3          3
                     Alternative Sum Formula
                               a1  a1r     n
                          Sn 
                                 1 r
                                     n 1
      We know that:         an  a1r
                                     n 1
      Multiply by r:      an  r  a1r  r
      Simplify:             an r  a1r n


                                a1  an r
      Substitute:          Sn 
                                  1 r
Jeff Bivin -- LZHS
                     Find the sum of the
                      geometric Series
  a1  5                         a1  an r
                            Sn 
       256                         1 r
  an                              5 
                                       256  2 
       729                                   
                                       729  3 
        2                   Sn   
    r                                 1
                                          2
                                          3
        3
                          512
                     5                     10423
              Sn         2187
                          1
                                       Sn 
                          3
                                             729
Jeff Bivin -- LZHS
                      10

Evaluate             2
                     k 1
                            k
                                = 2 + 4 + 8+…+1024
                                      a1  a1r
                                             n
                                Sn 
          a1 = 2                        1 r
          r=2                        2  22   10
                                Sn 
          n = 10
                                        1 2
          an = 1024
                                     2  2  1024
                                Sn 
                                          1
                                      2046
                                Sn            2046
Jeff Bivin -- LZHS
                                       1
                     8

Evaluate  3  2 j 1 = 3 + 6 + 12 +…+ 384
                     j 1

                                  a1  a1r
                                         n
                            Sn 
          a1 = 3                    1 r
          r=2                    3  3 2  8
                            Sn 
          n=8
                                    1 2
          an = 384
                                 3  3  256
                            Sn 
                                      1
                                    765
                             Sn           765
Jeff Bivin -- LZHS
                                    1
                      Review -- Geometric

                     nth term        Sum of n terms
                                          a1  a1r   n
                                     Sn 
             an =       a1·r (n-1)          1 r
                                          a1  an r
                                     Sn 
                                            1 r

Jeff Bivin -- LZHS
                     Geometric

                     Infinite Series




Jeff Bivin -- LZHS
                       The Magic Flea                 There is
                     (magnified for easier viewing)   no flea
                                                       like a
                                                       Magic
                                                        Flea




Jeff Bivin -- LZHS
                             The Magic Flea
                       (magnified for easier viewing)




              ...    1 1
                     32 16
                             1
                             8
                                 1
                                 4
                                                 1
                                                 2



           1 1 1  1   1
       S           ...  1
           2 4 8 16 32
Jeff Bivin -- LZHS
                     Sum it up -- Infinity

                             
                                                   a1
                      S    a r
                             i 1
                                    1
                                         n 1
                                                
                                                  1 r

                             for        r 1


Jeff Bivin -- LZHS
           Remember --The Magic Flea
            1 1 1  1    1
       S             ...  1
            2 4 8 16 32
                           a1
                     S 
                          1 r
        a1  2
             1             1
                     S   2

         r 21            1 1
                             2
                                 1
                          S    2
                                 1
                                     1
                                 2
Jeff Bivin -- LZHS
                         2 2    2
             S  6  2         ...
                         3 9 27
                            a1
                      S 
          a1  6           1 r
               1
          r          S 
                            6
               3           1 1 3


                            6
                     S    2
                                    9
                            3
Jeff Bivin -- LZHS
                     A Bouncing Ball
  rebounds ½ of the distance from which it fell --
     What is the total vertical distance that the
     ball traveled before coming to rest if it fell
      from the top of a 128 feet tall building?




Jeff Bivin -- LZHS
                     A Bouncing Ball
      Downward = 128 + 64 + 32 + 16 + 8 + …
                 a1   128   128
           S             1  256
                1 r 1 12   2




Jeff Bivin -- LZHS
                     A Bouncing Ball
          Upward        = 64 + 32 + 16 + 8 + …
                   a1   64     64
             S             1  128
                  1 r 1 12    2




Jeff Bivin -- LZHS
                     A Bouncing Ball
Downward = 128 + 64 + 32 + 16 + 8 + … = 256
  Upward =       64 + 32 + 16 + 8 + … = 128
                              TOTAL = 384 ft.




Jeff Bivin -- LZHS
                     A Bouncing Ball
  rebounds 3/5 of the distance from which it fell --
         What is the total vertical distance that the ball
      traveled before coming to rest if it fell from the top
                   of a 625 feet tall building?




Jeff Bivin -- LZHS
                     A Bouncing Ball
Downward = 625 + 375 + 225 + 135 + 81 + …
            a1   625   625
      S             2  1562 .5
           1 r 1 53
                        5




Jeff Bivin -- LZHS
                     A Bouncing Ball
          Upward       = 375 + 225 + 135 + 81 + …
              a1   375   375
        S             2  937 .5
             1 r 1 53
                          5




Jeff Bivin -- LZHS
                     A Bouncing Ball
Downward = 625 + 375 + 225 + 135 + 81 + … = 1562.5
  Upward =         375 + 225 + 135 + 81 + … = 937.5

                                  TOTAL = 2500 ft.




Jeff Bivin -- LZHS
                Find the sum of the series
  S n  .9  .09  .009  .0009  .00009  ...
                                 a1
              a1  .9     S 
                               1 r
                r  .1           .9
                          S 
                               1  .1
                                .9
                          S         1
                                .9
Jeff Bivin -- LZHS
                     Fractions - Decimals
                     .1    1
                            9           .6    6
                                               9      2
                                                       3

                     .2    2

                                        .7 
                            9                  7

                     .3        
                            3       1          9
                            9       3

                     .4    4           .8    8
                                               9
                            9

                     .5    5
                            9
                                        .9    9
                                               9   1

Jeff Bivin -- LZHS
                     Let’s try again
                               1
                        .3 
                               3
                               1
                        .3 
                               3
                               1
                        .3 
                               3
                               3
                        .9      1
                               3
Jeff Bivin -- LZHS
                       One more
                      10 x  9. 9
               let       x  .9
           subtract    9x  9
                        x  9 1
                            9




Jeff Bivin -- LZHS
                     OK now a series
  . 9  .9  .09  .009  .0009  .00009  ...
                                a1
                         S 
         a1  .9              1 r
           r  .1               .9
                         S 
                              1  .1

                                .9
                     . 9  S     1
                                .9
Jeff Bivin -- LZHS
                     .9 = 1
                                              .9 = 1
                                 That’s All
                                  Folks




                              .9  1
Jeff Bivin -- LZHS

				
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