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 10-15-2012, 01:13 PM #1 Junior Member   Join Date: May 2012 Posts: 49 Math Rules What was the equation for the sum of all terms in a arithmetic and geomatric sequence? For example i saw question that stated find the sum of the first 50 terms in this sequence. I know it has a sigma bit couldn't, for the life of me, remember what it was. Reply
 10-15-2012, 01:40 PM #2 Junior Member   Join Date: Aug 2012 Posts: 59 I just remember the formula of the sum of arithmatic sequence: S= [(the first number + the final number)* the number of numbers]/2 Or the first number + (n-1)*distance (n-1: the n-1 th term in the sequence) Reply
 10-15-2012, 01:52 PM #3 Member   Join Date: Mar 2011 Location: Staten Island, NY Posts: 999 The following little trick for an arithmetic series is better than memorizing a formula. See if you can explain what I'm doing: 1 + 2 + 3 + ... + 50 50 + 49 + 48 + ... + 1 51 + 51 + 51 + ... + 51 = 50(51) = 2550. Finally divide by 2: 2550/2 = 1275. If you want to know the formula, it is most easily remembered as (number of terms)*(average of first and last term) = 50(1 + 50)/2 = 1275. I haven't yet seen problem on the SAT where you need the arithmetic series formula (but there are often "differences of large sums" problems) Reply
 10-15-2012, 02:13 PM #4 Junior Member   Join Date: May 2012 Posts: 49 Thanks DrSteve But do you know the rule for the geomatric series formula? Reply
 10-15-2012, 02:15 PM #5 Junior Member   Join Date: Jan 2012 Posts: 231 Also, Dr Steve, are there any other cool formulas we should know? Reply
 10-15-2012, 02:52 PM #6 Member   Join Date: Mar 2011 Location: Staten Island, NY Posts: 999 I have given a full list of "cool" formulas in my article "The Math Formulas You Should Memorize for the SAT." I've posted it on this forum several times - just do a search. The geometric series formula is not one of these - there is no need to memorize this one. That said, here it is: Gn = g[(1-r^n)/(1-r)] where g is the first term of the series, r is the common ratio, and n is the number of terms. The infinite geometric series has a simpler formula: G = g/(1-r) if -11, then the infinite geometric series has no sum. Reply
 10-15-2012, 04:38 PM #7 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 A simple proof for a geometric series: Let S = 1 + r + r^2 + r^3 + ... (assuming S converges) Then rS = r + r^2 + r^3 + ... = S - 1 Solving for S gives S = 1/(1-r) (where |r| < 1). Reply
 10-16-2012, 09:08 AM #8 Junior Member   Join Date: Sep 2012 Posts: 88 DrSteve, I tried to employ your approach to the following question_ which is by the way from the Test on January 2011_ but it didn't work. I'm sure there was a flaw in my calculation or something. So the question is: The first three terms of a sequence are given: 1/(1)(2), 1/(2)(3), 1/(3)(4) The nth term of the sequence is 1/(n)(n+1), which is equal to 1/n - 1/n+1. What is the sum of the first 50 terms of this sequence? Please explain it to me. Reply
 10-16-2012, 12:00 PM #9 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 @Afaloo, the sequence is not arithmetic. Clearly, 1/(n*(n+1)) = 1/n - 1/(n+1). The sum of the first 50 terms is S = 1/(1*2) + 1/(2*3) + ... + 1/(50*51), which is equal to S = (1/1 - 1/2) + (1/2 - 1/3) + ... + (1/50 - 1/51) Here, notice that everything cancels out (-1/2 cancels out 1/2, etc.) leaving 1/1 - 1/51, or 50/51. This is known as a telescoping series. Reply
 10-16-2012, 12:27 PM #10 Junior Member   Join Date: Sep 2012 Posts: 88 @rspence, wow! Thank you so much, I really appreciate your help. I will post a couple of questions in a new thread, so I will be glad if you go through them once you have the time. Reply
 10-16-2012, 01:17 PM #11 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 Sure, no problem. Reply

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