<p>the first three terms of a sequence are:</p>
<p>1/(1)(2) , 1/(2)(3) , 1/(3)(4)</p>
<p>the nth term of a 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 the sequence?</p>
<p>a) 1
b) 50/51
c) 49/50
d) 24/50
e) 1/(50)(51)</p>
<p>This is a tough question. Anyone got the answer?</p>
<p>btw this qestin came in jan 2011 SAT!</p>
<p>
This is not true. 1/n - 1/(n-1)= -1/n(n-1) which is not equal to 1/n(n+1)</p>
<p>Can you check the question again ?</p>
<p>in fact it should be 1/n(n+1)= 1/n - 1/(n+1)
therefore the solution comes easy because adding the terms eliminates most of them:
the sum comes as:
1/1 -1/2 +(1/2 -1/3) + (1/3- 1/4) +…(1/50 - 1/51)= 1/1 -1/51= 50/51</p>
<p>oo sorry yes there’s a correction --</p>
<p>its 1/n - 1/(n+1)</p>