An algebra problem by Sean Ty

Algebra Level 5

Evaluate: 2013 + 276 2027 + 278 2041 + 280 2055 + . . . \displaystyle \sqrt{2013+276\sqrt{2027+278\sqrt{2041+280\sqrt{2055+...}}}}

This problem is not original.


The answer is 285.

Sean Ty by Sean Ty , 21, Philippines

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2 solutions

Using Ramanujan teorem :

( x + n + a ) \displaystyle (x+n+a) = a x + ( n + a ) 2 + x a ( x + n ) + ( n + a ) 2 + ( x + n ) . . . . \displaystyle \sqrt{ax+(n+a)^{2}+x\sqrt {a(x+n)+(n+a)^{2}+(x+n)\sqrt{....}}}

you will get x=276, n=2, a=7 So, 2013 + 276 2027 + 278 2041 + 280 2055 + . . . \displaystyle \sqrt{2013+276\sqrt{2027+278\sqrt{2041+280\sqrt{2055+...}}}} = x+n+a=276+2+7=285

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Sidharth Ghoshal
Nov 29, 2014

http://vixra.org/pdf/1310.0177v1.pdf

Can be used to generate solution

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