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 10-18-2012, 11:45 AM #1 New Member   Join Date: Jan 2012 Posts: 13 Hard questions for 2000s and up First Question 1/(1)(2) , 1/(2)(3) , 1/(3)(4) The first three terms of a sequence are given. The nth term of the sequence is 1/(n)(n+1). Which is equal to 1/(n)-(n-1). What is the sum of the first 50 terms of this sequence? a) 1 b) 50/51 c) 49/50 d) 24/50 e) 1/(50)(51) Second Question Y=-2(x-2)^2 +3 In the xy-plane, line L passes through the point (4,-5) and the vertex of the parabola with the equation above. What is the slope of line L ? A) -4 b) -1/4 c) 0 d) 1/4 e) 4 Please I need the answers with persuasive explanation Reply
 10-18-2012, 03:54 PM #2 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 1. The sum of the first 50 terms is (1 - 1/2) + (1/2 - 1/3) + ... + (1/50 - 1/51) Every term except 1 and -1/51 cancels (this is known as "telescoping"). The sum is 1 - 1/51 = 50/51. 2. Check out this thread: Math Problems Reply
 10-22-2012, 04:44 AM #3 New Member   Join Date: Jan 2012 Posts: 13 I'm very thankful to you But I have a question about the firsr solution» how have we come with 1. And -1\51 Reply
 10-22-2012, 09:36 AM #4 Senior Member   Join Date: Mar 2012 Location: Cambridge, MA Posts: 1,913 First term is 1 - 1/2 (or 1/1 - 1/2). The 50th term is 1/50 - 1/51. When you telescope, all of the intermediate terms (1/2, 1/3, ..., 1/50) cancel out, leaving 1 - 1/51. Reply

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