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I have a sequence knowing that $a_1=15$, $a_{n+1} = a_n + \sqrt{2a_{n+1}-a_n}$. I want to find some the first terms. I tried by my hand

k := 15;
Solve[x == k + Sqrt[2 x - k], x]

{{x -> 20}}

k := 20;
    Solve[x == k + Sqrt[2 x - k], x]

{{x -> 21 + Sqrt[21]}}

I am trying to find a general formulas. I tried

ClearAll["Global`*"];
RSolve[{a[x, n + 1] == a[x, n]  + Sqrt[ 2 a[x, n + 1] - a[x, n] ], 
   a[x, 1] == 15}, a[x, n], n] // FullSimplify

{{a[x, n] -> (-6 + n) (-4 + n)}, {a[x, n] -> (2 + n) (4 + n)}}

The out put, e.g $(2 + n) (4 + n)$ is incorrect, I think that.

How to find some first terms of the sequence $a_1=15$, $a_{n+1} = a_n + \sqrt{2a_{n+1}-a_n}$?

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1 Answer 1

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Try this to give you some of the first terms.

sol=15;
Table[sol=b/.ToRules[FullSimplify[Reduce[{a==sol,b==a+Sqrt[2b-a]},b]]],{4}]

which returns

{20,
 21+Sqrt[21],
 22+Sqrt[21]+Sqrt[22+Sqrt[21]], 
 23+Sqrt[21]+Sqrt[22+Sqrt[21]]+Sqrt[23+Sqrt[21]+Sqrt[22+Sqrt[21]]]
}

If you want a decimal approximation of that then follow it with

N[%]

and get

{20.,25.5826,31.7384,38.4602}

Test that brutally. Is it exactly correct?

That does not give you a closed form algebraic solution but it does give you "some first terms of your sequence." Remove the FullSimplify to make it somewhat faster, but then you won't see the structure of the solutions.

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