Question

Let \(k=\sqrt{n+\sqrt{n+\sqrt{n}+\sqrt{n}+\sqrt{n}+ ~...}}\), \(n \in \mathbb{N}\)

Show that \(k\) only tends to a whole number only if \(n\) is the product of two consecutif whole number.

This square root thing is troubling me, it goes to infinity, how can I express it under a kind of series, do I have to use the telescopic method?

Answer 1

@TuringTest

Answer 2

just notice
\[k^2=n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n}+...}}}\]

Answer 3

So basically I send all n and square root to k's side, expand and simplify?

Answer 4

sorry my english is not well

Answer 5

what is the equivalent of second term in the right hand of equation i posted for u?

Answer 6

:| No idea D:

Answer 7

\[k^2=n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n}+...}}}=n+k\]

Answer 8

\(n+k=k^2\)? o.o that doesn't make sense D:

Answer 9

then n=k(k-1)
i think that is what u want

Answer 10

Where does n=k(k-1) come from?

Answer 11

Ohh, ok ok

Answer 12

http://puu.sh/Ejof

Like this?

Answer 13

thats right

Answer 14

So k and k-1 are two consecutive whole numbers, n is the product of two consecutive whole numbers and a natural number, therefore k is a whole number?

Answer 15

yes

Answer 16

Great! Thank you so much! :D

Answer 17

your welcome