Question

Just a little challenge problem.

If \(a_1, a_2, a_3 \cdots a_n\) are in arithmetic progression, show that\[\dfrac{1}{\sqrt{a_1} + \sqrt{a_2}} + \dfrac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \dfrac{1}{\sqrt{a_{n-1} } + \sqrt{a_{n}}} = \dfrac{n -1 }{\sqrt{a_1} + \sqrt{a_n}}\]

Answer 1

@amistre64 @ganeshie8 @phi

Answer 2

Ooh ~

Answer 3

ugh .... does putting everything in terms of a1 help any?

Answer 4

probably yes, lol.

Answer 5

what is the GCD of it all?

Answer 6

oh wait, do you have the solution to this?

Answer 7

of course not lol

Answer 8

\[(\sqrt{a_1}+\sqrt{a_2})(\sqrt{a_2}+\sqrt{a_3})\]
\[\sqrt{a_1a_2}+\sqrt{a_1a_3}+a_2+\sqrt{a_2a_3}\]
\[(\sqrt{a_1a_2}+\sqrt{a_1a_3}+a_2+\sqrt{a_2a_3})(\sqrt{a_3}+\sqrt{a_4})\]
\[\sqrt{a_1a_2a_3}+a_3\sqrt{a_1}+a_2\sqrt{a_3}+a_3\sqrt{a_2}+\sqrt{a_1a_2a_4}+\sqrt{a_1a_3a_4}+a_2\sqrt{a_4}+\sqrt{a_2a_3a_4}\]
not sure how practical this idea is

Answer 9

if we rationalis thhe denominators maybe?

Answer 10

anything is practical as long as you generalize it till \(n\) terms. :)

Answer 11

yes!

Answer 12

\[\dfrac{\sqrt{a_1} - \sqrt{a_2}}{a_1 - a_2} + \dfrac{\sqrt{a_2} - \sqrt{a_3}}{a_2-a_3} + \cdots + \dfrac{\sqrt{a_{n-1}} - \sqrt{a_n}}{a_{n-1} - a_n} = \dfrac{(n -1)(\sqrt{a_1} - \sqrt{a_n}) }{a_1-a_n}\]

Answer 13

now we may be able to split the fractions into single denominators and it might ... just might .... telescope?

Answer 14

single numerators lol

Answer 15

Hah, yup.

Answer 16

On the LHS, we have\[\dfrac{\sqrt{a_1} - \sqrt{a_n}}{-d}\]and the same for the RHS.

Thanks, good proof!

Answer 17

ill take your word for that ;)