Sum of a Sequence

Date: 06/10/99 at 14:25:20
From: Bethany
Subject: Sum of an embedded arithmatic sequence

Our sequence is (3, 4, 6, 9, 13, ..., 499503). We know that there are 
1000 terms in the sequence. How would we figure out a formula for the 
sum of this sequence? We used Gauss' technique to find the number of 
terms by plugging in like this: 3+(n-1)(n-1+1)/2 = 499503.  We got 
1000 terms. We know this is correct. How would we find the sum of the 
sequence?

Date: 06/10/99 at 16:45:00
From: Doctor Anthony
Subject: Re: Sum of an embedded arithmatic sequence

Make up a difference table

   n =  1     2     3    4      5      6      7 ........
f(n) =  3     4     6    9     13     18     24 ........

1st Diff   1     2     3    4      5      6
2nd Diff      1     1     1     1      1

If the second differences are constant, the nth term will be a 
quadratic expression in n.

So we assume    f(n) =  an^2 + bn + c

    n=1                 a +  b +  c  =  3
    n=2                4a + 2b +  c  =  4
    n=3                9a + 3b +  c  =  6

So we have 3 equations with 3 unknowns a, b, and c.

The solutions are   a = 1/2,  b = -1/2,  c = 3

And therefore   f(n) = (1/2)n^2 - (1/2)n + 3

Check with n = 5: f(n) = 12.5 - 2.5 + 3 = 13, which checks. 

n = 1000  so we get: f(n) = 500000 - 500 + 3 = 499503, which also 
checks.

So we must sum the following

   1000
    SUM[n^2/2 - n/2 + 3]
    n=1

and using the standard formulae for SUM(n^2) and SUM(n) we get

  (1/2)n(n+1)(2n+1)/6 - (1/2)n(n+1)/2 + 3n

     n(n+1)[2n+1 - 3]
   = ----------------  + 3n
            12

     n(n+1)(2n-2)
   = ------------ + 3n
          12

     n(n+1)(n-1)           n(n^2 - 1) + 18n
   = ----------- + 3n   =  ----------------
         6                       6

Check when n = 3. The sum of the first 3 terms should be 13

     3 x (9 - 1) + 54
     ----------------  =  13  so our formula is correct.
             6

Finally, put n = 1000 into the formula

     1000(1000^2 - 1) + 18000
     ------------------------  =  166669500
                6

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/