Thaddeus Function

We can rationalize the denominator to get $\frac{\sqrt{2}- \sqrt{1}}{2-1}+\frac{\sqrt{3}- \sqrt{2}}{3-2} + \cdots + \frac{\sqrt{n+1}- \sqrt{n}}{n+1-n} =$ $(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n+1} - \sqrt{n})$ We see that all the terms from $\sqrt{2}$ to $\sqrt{n}$ will cancel out (this is called a telescoping series.). Thus we will end up with $\sqrt{n+1} - \sqrt{1}.$ Now we just have to make sure $n+1$ is a square and we will be finished. Since there are 31 positive perfect squares less than or equal to 1000, we can let $n+1$ be these squares. However, we cannot let it be 1 or $n$ would equal 0 so our answer is 30

Rationalizing denominators using conjugates, the function becomes f(n)= $\frac {\sqrt{1}-\sqrt{2}}{-1} + \frac {\sqrt{2}-\sqrt{3}}{-1} \dots +\frac {\sqrt{n}-\sqrt{n+1}}{-1}$ . Adding yields f(n)= $\frac {1-\sqrt{n+1}}{-1}$ , or $\sqrt{n+1} - 1$ . For $\sqrt{n+1} - 1$ to be an integer, $\sqrt{n+1}$ must be an integer. Hence, $(n + 1)$ must be a perfect square. There are 31 perfect squares between 1 and 1000 (1,4,9 ... ,961), but because n can't be 0, there are $30$ integers n so that f(n) is an integer.

1/(√1+√(2 )) may be written as

(√(1 )-√2)/(1-2) in it' s rationalized form,

similarly,

1/(√2+√(3 )) may be written as (√(2 )-√3)/(2-3)

Applying this generalization throughout, upto 'n' terms,

we have, f(n) = { (√1-√2)+(√2-√3)+(√3-√4)+⋯+(√n-√(n+1)) } / (-1)

which may be simplified as,

f(n) = (√(n+1)) - 1

now, The limit for integer 'n' given in the problem range from 1 to 1000. and since we need only integers, which implies that the values for √(n+1) must be integers as well. so, the numbers whose squares less than 1000 must be chosen (exclude zero, as nЄ [1, 1000] ).

Clearly there are 30 numbers which satisfy this condition, viz; 2^2, 3^2... 31^2. so the number of integers is 30 >>Ans

Any expression of the form 1/ ( √n + √n+1 ) may be expressed as: [(√n+1)^2 - (√n)^2 ] / ( √n + √n+1 ) =[( √n + √n+1 )*( √n - √n+1 )] / ( √n + √n+1 ) =( √n - √n+1 ).

Hence 1/√1 + √2 = √2 - √1 & 1/√3 + 1/√2=√3 - √2

For any n say 3 we can reduce the expression to : √2 - √1 + √3 - √2 + √4 - √3. Crossing the diagonal terms, we have √4 - √1 =3. For n=4, we are left with √5 - √1 : an irrational identity, hence discarded.

Our solution set would hence be (all the perfect squares - 1) i.e. 3, 8, 15, 24, 35, 48...... 960 (31^2 =961). And n=1 is not a valid solution.

Hence the required answer is 31-1= 30

multiply & divide by sqrt(n+1) - sqrt(n) with each term of the function. after simplification the function will be... f(n) = sqrt( n+1) - 1 Now, the function to be an integer (n+1) should be a PERFECT SQUARE. since, 1<= n<=1000, 2<=n+1<=1000 and n+1 can take the values 2^2 (4) , 3^2 , 4^2 ............upto 31^2 (961) therefore total number of values of n is 30.

We have f(n) = \frac {1}{\sqrt{1} + \sqrt{2}} + \frac {1}{\sqrt{2} + \sqrt{3}} + \ldots + \frac {1}{\sqrt{n} + \sqrt{n+1}} = \frac {\sqrt{2} - \sqrt{1}}{(\sqrt{1} + \sqrt{2}) \times (\sqrt{2} - \sqrt{1})} + \frac {\sqrt{3} - \sqrt{2}}{(\sqrt{2} + \sqrt{3}) \times (\sqrt{3} - \sqrt{2})} + \ldots + \frac {\sqrt{n+1} - \sqrt{n}}{(\sqrt{n} + \sqrt{n+1}) \times (\sqrt{n+1} - \sqrt{n})} = \frac {\sqrt{2} - \sqrt{1}}{2 - 1} + \frac {\sqrt{3} - \sqrt{2}}{3 - 2} + \ldots + \frac {\sqrt{n+1} - \sqrt{n}}{n+1 - n} = \sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \ldots + \sqrt{n+1} - \sqrt{n} = \sqrt{n+1} - 1 So, in order to make f(n) be an integer \Rightarrow \sqrt{n+1} must be an integer \Rightarrow n+1 \in {1^2 ; 2^2 ; ... ; 31^2}(because n \leq 1000 \Rightarrow n+1 \leq 1001) But if n+1 = 1 \Rightarrow n=0 ( not be valid) Thus, there are 30 positive intengers n .

First we need to show that 1/(sqrt(n+1)-sqrt(n))=sqrt(n+1)-sqrt(n)

1/(sqrt(n+1)+sqrt(n))=(sqrt(n+1)-sqrt(n))/((sqrt(n+1)-sqrt(n))*(sqrt(n+1)+sqrt(n)))=(sqrt(n+1)-sqrt(n))/(n+1-n)=sqrt(n+1)-sqrt(n)

Using this identity we get f(n)=(sqrt(2)-sqrt(1)) + (sqrt(3)-sqrt(2) ) + ........ + (sqrt(n)-sqrt(n-1)) + (sqrt(n+1)-sqrt(n)) Almost everything cancels we get f(n)=-sqrt(1)+sqrt(n+1)=sqrt(n+1)-1 f(n) is an integer => f(n) is positive since n is a positive integer we dont have to worry about that because sqrt(n+1) is greater than or equals 1 for every positive integer n f(n) i an integer => sqrt(n+1) is an integer => there exists a positive integer k such that n+1=k² => since 1≤n≤1000 then 2≤k≤31 => k is an integer then k is within the set {2,3,__,31} hence k has 30 possibilities and because n=k²-1 then n has 30 possible values

For the f(n) to be an integer the denominator of the sum of the fractions must be an integer. Firstly we notice that when two fractions are simplified all the irrational square roots in the denominators besides the two at each end are rationalized because they are multiplied by them selves. For example: \frac {1}{\sqrt{1} + \sqrt{2}} + \frac {1}{\sqrt{2} + \sqrt{3}} = -1 + \sqrt{3} Here we see that the irrational number \sqrt{2} in the middle has been rationalized, but the two at the ends have not. We notice two things, firstly that \sqrt{1} just stays as a rational so we do not have to worry about it and that if the number on the end were to be a square number (e.g. 4) then the value of f(n) would become an integer. For example: \frac {1}{\sqrt{1} + \sqrt{2}} + \frac {1}{\sqrt{2} + \sqrt{3}} + \frac {1}{\sqrt{3} +\sqrt{4}} = 1 This shows that if the last irrational in f(n) is a square root of a square (e.g. 4) then f(n) will be an integer. Since the last square root in the f(n) is \sqrt{n+1} then to find all the n values that f(n) would give an integer we would just have to find all the number 1 \ leq n \ leq 1000 that is 1 less then a square number. There are 31 square numbers under 1000 and above or equal to 1, because 31^2= 961 and 32^2= 1024 so 31 is the square root of the smallest square number under 1000. This makes the number of square roots under 1000 and above or equal to 31, but we do not count 1 because n having to be bigger or equal to 1 cannot allow n+1 to be equal to 1.