24i have solved the ques..but ans isnt matching..
plz post ur solns...
For a sufficiently large value of n the sum of the square roots of the first n positive integers is approximately equal to?
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11 Answers
24
62What are the options..
I mean this cannot be answered without options :D
It is very close to a lot of numbers :P
62okie.. if that is what you meant,
asymptotically we can say that the integral will be a+b.n3/2
Now the difference will be equal to b{(n+1)3/2-n3/2) = √n+1
so b is the limit as n goes to infinity of √n+1/ {(n+1)3/2-n3/2)
√n+1/ {(n+1)3/2-n3/2)
Now i guess you can complete this one..
62The b part is easy to find and it will indeed be equal to 2/3
what I am not able to eliminate right now is a
here in my solution a turns out to be zero.. (according to the answer) but I dont have a proof for it right now..
my solution seems to give a+2/3 n3/2
I guess I am missing out something very very simple :(
24here is the soln given
\lim_{n\rightarrow \infty}\frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n}}{n\sqrt{n}}=\int_{0}^{1}{\sqrt{x}}dx=\frac{2}{3}
SoS_n=\frac{2n^{3/2}}{3}
62oh oh...
This is simple and awesome :)
The question is same as finding the sum on the left and recognizing that it is finite..
The integral comes from the definition of integration....
And the RHS is the value of the integral..
Hence we know that limiting case the integral is equal to this sum
and hence the answer...
Did you understand it now eureka?
24SIr, I understod that first line..but not the second line ...
Sn=2n3/23
62sn is the numerator.. the one whose limit we have to find..
so we know that sn/(n3/2) tends to 2/3
hence sn tends to 2/3 x n3/2