The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1. What is the sum of the first 60 terms of this sequence?	Pick: A) (1+1/2)/ (1/51 +1/ 52) B) (1+1/2) – (1/61 +1/ 62) C) (1+1/2)* (1/51 +1/ 52) D) (1+1/2) + (1/51 +1/ 52) E) (1/51 +1/ 52)
The answer would most certainly be[B]. But the question needs a slight modification.n&gt;=1, since the answer does consider a1 under the sum.   The sequence is :   a1 = 1-1/3   a2 = 1/2 - 1/4   a3 = 1/3 - 1/5....   We can observe that the third term in the sequence cancels the negative term in the first. A similar approach can be seen on all the terms and we would be left with 1 + 1/2 from a1 and a2 along with -1/62 and -1/61 from a60 and a59 term which could not be cancelled.   Hence the sum = (1+1/2) – (1/61 +1/ 62)   Answer : (1+1/2) – (1/61 +1/ 62)
<result>B</result>
The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1. What is the sum of the first 60 terms of this sequence?