Parth (parthkohli):

Just a little challenge problem. If $$a_1, a_2, a_3 \cdots a_n$$ are in arithmetic progression, show that$\dfrac{1}{\sqrt{a_1} + \sqrt{a_2}} + \dfrac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \dfrac{1}{\sqrt{a_{n-1} } + \sqrt{a_{n}}} = \dfrac{n -1 }{\sqrt{a_1} + \sqrt{a_n}}$

4 years ago
OpenStudy (igreen):

@amistre64 @ganeshie8 @phi

4 years ago
sammixboo (sammixboo):

Ooh ~

4 years ago
OpenStudy (amistre64):

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

4 years ago
Parth (parthkohli):

probably yes, lol.

4 years ago
OpenStudy (amistre64):

what is the GCD of it all?

4 years ago
Parth (parthkohli):

oh wait, do you have the solution to this?

4 years ago
OpenStudy (amistre64):

of course not lol

4 years ago
OpenStudy (amistre64):

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

4 years ago
OpenStudy (amistre64):

if we rationalis thhe denominators maybe?

4 years ago
Parth (parthkohli):

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

4 years ago
Parth (parthkohli):

yes!

4 years ago
OpenStudy (amistre64):

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

4 years ago
OpenStudy (amistre64):

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

4 years ago
OpenStudy (amistre64):

single numerators lol

4 years ago
Parth (parthkohli):

Hah, yup.

4 years ago
Parth (parthkohli):

On the LHS, we have$\dfrac{\sqrt{a_1} - \sqrt{a_n}}{-d}$and the same for the RHS. Thanks, good proof!

4 years ago
OpenStudy (amistre64):

ill take your word for that ;)

4 years ago