Let f(n)=1/√1+√2+1/√2+√3+1/√3+√4…1/√n+√n+1. For how many positive integers n, in the range 1≤n≤1000, is f(n) an integer?
To find the values of n for which f(n) is an integer, we need to examine the expression and look for patterns.
The given expression can be written as:
f(n) = 1/√1+√2 + 1/√2+√3 + 1/√3+√4 + ... + 1/√n+√(n+1)
We can simplify this expression by rationalizing the denominators:
f(n) = (1/√1+√2) + (1/√2+√3) + (1/√3+√4) + ... + (1/√n+√(n+1))
= (√1-√2) + (√2-√3) + (√3-√4) + ... + (√n-√(n+1))
Notice that many terms cancel each other out:
(√2-√2), (√3-√3), (√4-√4), etc. all equal 0.
We can rewrite the expression with only the first and last non-cancelled terms:
f(n) = (√1-√(n+1))
Now, for f(n) to be an integer, (√1-√(n+1)) must be an integer, which means (√(n+1)) must be an integer.
If (√(n+1)) is an integer, then (n+1) must be a perfect square.
Let's analyze the range of values for n: 1≤n≤1000.
To find the number of perfect squares within this range, we can calculate the square root of the minimum and maximum values of n+1:
√2 ≈ 1.41
√1001 ≈ 31.64
There are 31 perfect squares between 1 and 1000.
Therefore, there are 31 positive integers n, in the range 1≤n≤1000, where f(n) is an integer.