Sum of a SequenceDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/