<p>I’ll do question 2:</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>On hard problems - and this one qualifies - the SAT will rarely give you information that’s not useful, so the fact that 1/(n)(n +1) is equal to (1/n) - (1/n + 1) is a good route to try. Also, on almost all sequence questions, writing out the terms let’s you see a pattern. In this case, write out the first three terms and the last term in the form of (1/n) - (1/n + 1):</p>
<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 add them together:
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + . . . + 1/50 - 1/51</p>
<p>All the middle terms are going to cancel out and you’ll be left with the first and last term: </p>
<p>1/1 - 1/51</p>
<p>So the answer is 50/51</p>