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 80 terms of this sequence? Select.
A) (1/81 +1/ 82)  B) (1+1/2) + (1/81 +1/ 82)  C) (1+1/2) / (1/81 +1/ 82)  D) (1+1/2) + (1/81 -1/ 82)  E) (1+1/2) – (1/81 +1/ 82)
The answer would most certainly be[E]. 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 -1/82 and -1/81 from a80 and a79 term which could not be cancelled.   Hence the sum = (1+1/2) – (1/81 +1/ 82)   Answer : (1+1/2) – (1/81 +1/ 82)
<result>E</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 80 terms of this sequence?