1you can take A=2011+2007 then subsequently other will be A+2 and so on
√A√A+2√A+4√A+6...∞
if anyone knows how to solve this further can do it
but i dont know if this method is correct or not....
evaluate-
\sqrt{2011+2007\sqrt{2012+2008\sqrt{2013+2009\sqrt{2014+.........\infty}}}}
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UP 0 DOWN 0 2 8
8 Answers
1
1ya as prophet sir said its generalisation of Ramanujan's problem.
actually prob which ramanujan posted in Journal of Indian Mathematical Society was this
a variant of prob asked
but no one cud solve it in stipulated time of six months !
then he only gave teh soln.
1Solution of f ( x ) ---
1 > Observe that , f ( x ) must be a polynomial of first degree , as in other cases , the powers of x
in f ( x ) won't match on L . H . S and on R . H . S .
So , f ( x ) = c x + b
Again , f ( 0 ) = n + a
So , b = n + a
So , f ( x ) = c x + n + a .
Now , plug this f ( x ) in the recurrence relation , from where c comes out to be 1 .