Down and Up Series
 
David W. Hansen
© 2003
 
          The set of beautiful expressions below are called Down and Up Series because of the down and
 
   up nature of the terms in each series.
 
                                               Series                                             Sum               Sum + 1
 
                                                    1                                          =         1                   2 = 1(2)
 
                                              2 + 1 + 2                                  =         5                  6 = 2(3)
 
                                       3 + 2 + 1 + 2 + 3                           =        11                12 = 3(4)
 
                               4 + 3 + 2 + 1 + 2 + 3 + 4                    =        19                20 = 4(5)
 
                        5 + 4 + 3 + 2 + 1 + 2 + 3 + 4 + 5           =       29                30 = 5(6)
 
 
          Here’s a dot picture of the 4th up and down series (which looks like a V).
                                                               
                                                                           .                       .
                                      .   .               .   .
                                      .   .   .       .   .   .
                                      .   .   .   .   .   .   .
                                                                                                                            
                                          4  +  3  + 2  +  1 +  2  +  3  + 4
 
 
          The sums of the series shown above; namely, 1, 5, 11, 19, and 29 do not appear to have any
 
   pattern. However, if we add 1 to each sum, we get the values shown in the third column above which
 
   do factor nicely.    The five equations above suggest to us that, in general,
 
                 n  + . . . +  3 + 2 + 1 + 2 + 3 +  . . . + n = n(n+1) – 1,  for any natural number n.            (1)
                                     
          Let’s try to prove (1).     Since 1  +  2  +  3  +  .  .  .  +  n  =  n(n+1)/2,  we get
 
                                                      n  +   .  .  .  +  3  +  2  +  1  +  2  +  3  +  .  .  .   +  n
 
     =                                              (n  +  .  .  .  +  3  +  2  +  1)  +  (1 + 2 + 3 + . . .  + n)  –  1
 
     =                                                                          n(n+1)/2   +   n(n+1)/2  –  1
 
     =                                                                                 2[n(n+1)/2]  –  1
 
     =                                                                                     n(n+1)  –  1
 
                                                                                       which proves (1).
 
     Alternate proof.   Let S1   =    1    +     2     +   .   .   .   +    n   +   (n – 1)  +   .   .   .   +     2     +   1
 
       and                            S2   =    n    +   (n-1) +   .   .   .    +   1     +      2      +   .   .   .   +  (n-1)  +   n.
 
      Then,                S1 + S2    =  (n+1) + (n+1) +  .   .   .   +  (n+1) +   (n+1)   +   .   .   .   + (n+1) + (n+1).
 
     Now, S1 has n terms going up from 1 to n and n-1 terms going down from n-1 to 1. Thus, it has a
 
     total of 2n-1 terms, as do S2 and S1 + S2. So, S1 + S2 = (2n-1)(n+1). Solving this, for S2, we get
 
     S2 = (2n-1)(n+1) – S1. Since S1 is an up and down series, S1 = n2. Thus, S2 = (2n-1)(n+1) - n2
 
     =  2n2  +  n – 1 – n2   =   n2 + n – 1   =   n(n+1) – 1, which proves (1).   
 
 
     Example 1. Find the down and up series whose sum is 89. The sum of a down and up series is 
 
     n(n+1) – 1,  so n(n+1) – 1 = 89,   n2 + n – 90 = 0,    (n – 9)(n + 10) = 0, and n = 9 (or - 10).  The
 
     series is then    9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 89.
 
 
     Example 2. Find the number of terms in a down and up series whose sum is 209. Since the sum is
 
     given by n(n+1) – 1, we have n(n+1) – 1 =  209,   n2 + n – 210 = 0, (n – 14)(n + 15) = 0, and 
 
     n = 14 (or - 15). As discussed in the alternate proof above, a down and up series has 2n – 1 terms. 
 
     Thus, our series has  2n – 1  =  2(14) – 1  =  27 terms. The series is  
 
                                       14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
 
                                        + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14