SAT Math Question: Series

Test Preparation SAT Preparation
1

<p>This is from the January 2011 test.</p>

<ol>
<li>1/(1)(2) , 1/(2)(3) , 1/(3)(4)</li>
</ol>

<p>The first three terms of a sequence are given above. The nth term of the sequence is given 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?</p>

<p>The formula for the sequence has a peculiar similarity to finding the sun of a certain special series that increases by one with each number: (n(n+1))/2. However, I haven't really been able to progress further than that. Can someone explain this problem to me?</p>

2

<p>This was answered in depth in this thread: <a href=“http://talk.collegeconfidential.com/sat-preparation/1180319-sat-math-help.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1180319-sat-math-help.html&lt;/a&gt;&lt;/p&gt;

3

<p>1st term: 1/1 - 1/2
2nd term: 1/2 - 1/3
3rd term: 1/3 - 1/4
…
50th term: 1/50 - 1/51</p>

<p>Now when we add these up, notice all the cancellations that take place (each rightmost term cancels with the next leftmost term - thus we are only left with the very first term, and very last term).</p>

<p>We wind up with 1/1 - 1/51 = 51/51 - 1/51 = 50/51</p>

<p>Remark: You can also just do the computation in your calculator instead of getting a common denominator.</p>

<p>This is known as a telescoping sum.</p>

4

<p>Check this older thread: </p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/1131967-another-math-problem-january-sat.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/1131967-another-math-problem-january-sat.html&lt;/a&gt;&lt;/p&gt;

5

<p>nevermind this</p>

6

<p>Thanks everyone for the enlightening replies and links! I now completely understand how to solve this problem.</p>