The sequence of numbers a1, a2, a3, ..., an is defined by an = 1/n - 1/(n+2) for each integer n >= 1...
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?
Answer/Solution
(1+1/2) – (1/61 +1/ 62)
Steps/Work
The answer would most certainly be[B]. But the question needs a slight modification.n>=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 : B
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 : B