<p>Hello,</p>
<p>As a follow-up to a similar topic thread, I have a question about another math problem from the January 2011 SAT. I believe it was the other "Level 5" problem and asked for the sum of the first 50 terms in the given sequence. Somehow, I have been unable to come up with the correct answer. Can somebody please outline the soluton to me, either with or without a calculator? Thanks in advance.</p>
<p>Best Wishes,</p>
<p>Dennis</p>
<ul>
<li><p>Type 1 (Arithmetic progression):
1st term = x, 2nd term = x + d, 3rd term = x + 2d, …, nth term = x + nd
=> S = nx + [1 + 2 + 3 + … + (n-1)]d
1 + 2 + 3 + … + (n-1) = n(n-1)/2
=> S = nx + n(n-1)d/2 (S = sum, x = 1st term, n = number of terms, d = common difference)</p></li>
<li><p>Type 2 (Geometric progression):
1st term = x, 2nd term = xq, 3rd term = x(q^2), …, nth term = x(q^(n-1)]
=> S = x[1 + q + q^2 + … + q^(n-1)] = xA
A = 1 + q + q^2 + … + q^(n-1)
qA = q + q^2 + q^3 + … + q^n
=> (q - 1)A = q^n - 1
=> A = (q^n - 1)/(q - 1)
=> S = xA = x(q^n - 1)/(q - 1) (S = sum, x = 1st term, n = number of terms, q = common ratio)</p></li>
</ul>
<p>hope that helps :D</p>
<p>withoutdoubt, isn’t the answer you gave the general “skeleton” of progression questions that require summation?</p>
<p>dcc1950, I’m terribly sorry that I can’t help you out, even though I had given the January SAT but I have no clue about the question you are referring to.</p>
<p>yes, it is. if you want to be able to answer any similar question, you’d better scrutinize it, absorb it and figure it out. believe me, i don’t provide more information than needed.</p>
<p>Hello,</p>
<p>Please accept my apologies for my failure to be more specific with this problem. I’m still new here and learning my way around.</p>
<p>I will try to recreate this problem the best I can:</p>
<p>19)</p>
<p>1/(1)(2) , 1/(2)(3) , 1/(3)(4)</p>
<p>The first three terms of a sequence are given above. 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?</p>
<p>(A) 1
(B) 50/51
(C) 49/50
(D) 24/50
(E) 1/(50)(51)</p>
<p>The correct answer is B. However, I can’t seem to get a handle on the solution.</p>
<p>Any assistance would be deeply appreciated.</p>
<p>Thanks!</p>
<p>Dennis</p>
<p>My attempt at explaining it on paper:</p>
<p><a href=“http://i184.■■■■■■■■■■■■■■■/albums/x153/spl10246/scan0001-1.jpg[/url]”>http://i184.■■■■■■■■■■■■■■■/albums/x153/spl10246/scan0001-1.jpg</a></p>
<p>But when I saw it on the SAT in January, I got scared and used my calculator to do it. (and got it right quickly :p)</p>
<p>how do you do something like this on the calculator?</p>
<p>Two possibles routes for the regular SAT student.</p>
<ol>
<li>The solution posted by OtherWindow is a good one. In this case, it is a good use of time to work on the pattern. One differenct approach is to focus on the successive elimination of terms in the pattern.</li>
</ol>
<p>1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 … -1/50 +1/50 - 1/51</p>
<p>At the end, you have 1 - 1/51 or 51/51 - 1/51 = 50/51</p>
<ol>
<li>This one is tricky and requires trusting ETS to stay true to their patterns. </li>
</ol>
<p>Take a look at the answers. </p>
<p>(A) 1
(B) 50/51
(C) 49/50
(D) 24/50
(E) 1/(50)(51)</p>
<p>This sequence starts with 1/2 and adds increasingly smaller fractions. While the limit will be 1, it cannot be smaller than 1/2 (because of the additions)</p>
<p>This means that 1, 24/50, and 1/50.51 are impossible answers. </p>
<p>Now we have two choices (B and C) that are very close to 1. One could guess randomly, and some courses advocate that tactic. However, one could also remember that ETS loves to give a “trick answer.” That answer is usually a derivation from the correct one. In this case, an educated guess is that 1/(50)(51) is a dead giveaway. So pick, B. </p>
<p>PS Advanced students have a number of additional options, but this is the SAT.</p>
<p>Hello OtherWindow and ziggi,</p>
<p>I would like to thank both of you for the comments and solutions you provided for this problem.</p>
<p>I understand it now.</p>
<p>I think the intimidation factor of this problem had gotten to me and disrupted my ability to approach this problem with a clear head.</p>
<p>Thanks again!</p>
<p>Best Wishes,</p>
<p>Dennis</p>
<p>What’s the exact problem? I would like to try doing it.</p>
<p>1/(1)(2) , 1/(2)(3) , 1/(3)(4)</p>
<p>The first three terms of a sequence are given above. 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?</p>
<p>(A) 1
(B) 50/51
(C) 49/50
(D) 24/50
(E) 1/(50)(51)</p>
<p>It is actually very obvious if you just fill in the numbers. Let’s do it with the first ones (to 3)
You get:
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 = 3/4. Do you see it? 
It says 1 - 1/(n + 1), or even easier: n/(n + 1). Fill in 50 and you get: 50 / 51.</p>
<p>“Do you see it?”</p>
<p>1 - 1/51 = 50/51 ( all the other terms cancel out )</p>
<p>I do, in fact your’s is the fastest one :)</p>
<p>wow dutch nice work… anyone care to verify?</p>
<p>
</p>
<p>No need to verify. Dutch’s solution of canceling equal terms is the same I posted here:</p>
<p>
</p>
<p>Oh i see… thanks… yeah that doesn’t seem very difficult at all once you notice that simple pattern.</p>
![http://i184.■■■■■■■■■■■■■■■/albums/x153/spl10246/scan0001-1.jpg[/url]](http://i184.photobucket.com/albums/x153/spl10246/scan0001-1.jpg%5B/url%5D){kind=link}
