Donnie84 wrote:
The sequence of numbers \(a_{1}, a_{2}, a_{3},\) …\(a_{n}\) is defined by \(a_{n} = \frac{1}{n} - \frac{1}{n+2}\) for each integer n=>1 . What is the sum of the first 20 terms of the sequence?
A. \((1+\frac{1}{2})-\frac{1}{20}\)
B. \((1+\frac{1}{2})-\frac{1}{21}+\frac{1}{22}\)
C. \(1-(\frac{1}{20}+\frac{1}{22})\)
D. \(1-\frac{1}{22}\)
E. \(1-\frac{1}{20}-\frac{1}{22}\)
\(a_{n} = \frac{1}{n} - \frac{1}{n+2}\)
Now, the positive values(n) start from 1 to 20 while negative values(n+2) start from 2 to 22. So, the positive values and negative values will cancel out each other from n as 2 till n as 20.
What remains after addition of first 20 terms is positive values of the term when n is 1 & 2 and negative values of the term when n is 21 and 22.
Sum = \(a_{n} = \frac{1}{1} +\frac{1}{2}- \frac{1}{21}-\frac{1}{22}\)
B option requires a change in sign ahead of 1/22
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