+0  
 
0
44
6
avatar+6 

For any positive integer n, let \(f\left(n\right)=\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}\) find the sum of f(1)+f(2)+...+ f(40)

 Jun 29, 2023
 #1
avatar
0

We can simplify the expression for f(n) as follows:

f(n) = (4n + sqrt(4n^2 - 1))/(sqrt(2n + 1) + sqrt(2n - 1)) = (4n + sqrt(2^2(2n - 1)^2 - 1)) / (sqrt(2n + 1) + sqrt(2n - 1)) = (4n + 2(2n - 1)) / (sqrt(2n + 1) + sqrt(2n - 1)) = n + 1 / (sqrt(2n + 1) + sqrt(2n - 1))

Now, let's evaluate f(1), f(2), ..., f(40):

f(1) = 2 + 1 / (sqrt(3) + sqrt(1)) = 3 / 2 f(2) = 4 + 1 / (sqrt(5) + sqrt(3)) = 5 / 2 f(3) = 6 + 1 / (sqrt(7) + sqrt(5)) = 7 / 2 ... f(40) = 81 + 1 / (sqrt(83) + sqrt(81)) = 82 / 2

 

The sum of f(1), f(2), ..., f(40) is:

3 + 5 + 7 + ... + 82 = \frac{3(1 + 2 + ... + 40)}{2} = \frac{3 \cdot 40 \cdot 41}{2 \cdot 2} = \frac{63 \cdot 41}{2} = \boxed{1263}

Therefore, the sum of f(1), f(2), ..., f(40) is 1263.

 Jun 29, 2023
 #3
avatar
0

Thanks, that's right!

Guest Jun 30, 2023
 #2
avatar+389 
+1

\(\displaystyle f(n) = \frac{4n+\sqrt{4n^{2}-1}}{\sqrt{2n+1}+\sqrt{2n-1}} \\ \displaystyle= \frac{4n+\sqrt{2n+1}.\sqrt{2n-1}} {\sqrt{2n+1}+\sqrt{2n-1}} .\frac{\sqrt{2n+1}-\sqrt{2n-1}}{\sqrt{2n+1}-\sqrt{2n-1}} \\ \displaystyle=\frac{4n\sqrt{2n+1}+(2n+1)\sqrt{2n-1}-4n\sqrt{2n-1}-(2n-1)\sqrt{2n+1}}{(2n+1)-(2n-1)}\\ \displaystyle=\frac{(2n+1)\sqrt{2n+1}-(2n-1)\sqrt{2n-1}}{2}.\)

 

\(n=1 : f(1)=(3\sqrt{3}-1\sqrt{1})/2, \\ n=2 : f(2)=(5\sqrt{5}-3\sqrt{3})/2, \\ n = 3 : f(3) = (7\sqrt{7}-5\sqrt{5})/2, \\ ............................ \\ ............................\\ n=39 : f(39) = (79\sqrt{79}-77\sqrt{77})/2\\ n = 40 :f(40) =(81\sqrt{81}-79\sqrt{79})/2. \)

 

\(\displaystyle f(1)+f(2)+\dots+f(40)=-\frac{1\sqrt{1}}{2}+\frac{81\sqrt{81}}{2}=364.\)

.
 Jun 30, 2023
 #5
avatar
0

364 is the correct answer !!

Guest Jun 30, 2023
 #6
avatar+6 
-2

im pretty sure 1263 is correct, i ran some calculations and i got something pretty close to 1263

 Jul 1, 2023

5 Online Users

avatar
avatar
avatar