(1/(√b+√a))+(1/(√d+√c))+...+(1/(√x+√y))=(n-1)/(√b+√y)?
b,a,d,c,...,x,y (they are n letters and n is Subset of set of natural numbers) are an Arithmetic sequence and all of them are Positive so prove this phrase that it`s right
(1/(√b+√a))+(1/(√d+√c))+...+(1/(√x+√y))=
=(n-1)/(√b+√y)
Update:b,a,d,c,...,x,y (they are n letters and n is Subset of set of natural numbers) are an Arithmetic sequence and all of them are Positive so prove this phrase that it`s right
(1/(√b+√a))+(1/(√d+√c))+...+(1/(√x+√y)
=(n-1)/(√b+√y)
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There are difficulties with the statement of the question!
Why are the letters not listed in alphabetical order, until "x,y"?
This is important, since, apparently, they are an arithmetic sequence.
Why does the next to last line end "...". It seems that it should just end.
When tried with 1,2,3,4, the stated equality failed, so I must ASSUME that the intent is:
There exists positive integers n, b,a,d,c,...x.y, such that b,a,d,c,...x,y are an arithmetic sequence and
1/(√b+√a)+1/(√d+√c)+...+1/(√x+√y)=(n-1)/(√b+√y).
I showed that it is not true for n=2 and 1,2,3,4. It is always true for n=1 and b,a!
I would be interested in any solution with n not 1.
I think this might be the real situation! All as I gave above, but equation is now the identity
1/(√b+√a)+1/(√a+√d)+1/(√d+√c)+1/(√c√e)...+1/(√x+√y)=(√b-√y)/(b-a).