Question

the square root of 1+\[1/(\sqrt{2}+1) +1/(\sqrt{3} +\sqrt{2}) + 1/(\sqrt{4} +\sqrt{3}) +............... 1/(\sqrt{324} +\sqrt{323}) is\]

Answer 1

options are 3root2 1/root2 2root3

Answer 2

the answer is 3root2

Answer 3

do you mean square root of this whole term written in long?

Answer 4

yes we have to find the square root

Answer 5

i think there is some trick.........

Answer 6

yes it is i am giving u the hints.. rationalize the each terms..!

Answer 7

the square root of 1+\[1/(\sqrt{2}+1) +1/(\sqrt{3} +\sqrt{2}) + 1/(\sqrt{4} +\sqrt{3}) +............... 1/(\sqrt{324} +\sqrt{323}) is\]

Answer 8

this is the question plzzz look at it

Answer 9

he square root of 1+
1/(2√+1)+1/(3√+2√)+1/(4√+3√)+...............1/(324−−−√+323−−−√)is

Answer 10

dont forgot to add that 1

Answer 11

do you know what a conjugate of a number is?

Answer 12

\[\frac{1}{a+b}*\frac{a-b}{a-b}=\frac{a-b}{a^2-b^2}\]

Answer 13

rationalizing the terms
in denomiantor of each terms u will get 1.. & in numerator the terms like this..
\[\sqrt{2}-1+\sqrt{3}-\sqrt{2}+.........................\sqrt{324}-\sqrt{323}\]

so each terms get cancelled and we are left with
\[\sqrt{324}\]

Answer 14

that called a telescoping sum right ?

Answer 15

take the square root of sqrt(324) that is square root of 324 which is 3root2

Answer 16

oh thanzzzz u really is a worth for me 1000000 thanzzzz