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Mathematics 17 Online
OpenStudy (anonymous):

The value of _____

hartnn (hartnn):

is _______

OpenStudy (anonymous):

\[\sqrt{1 + 2008\sqrt{1 + 2009\sqrt{1 + 2010\sqrt{1 + 2011.2013}}}}\]

OpenStudy (anonymous):

what is the value of the above ???

OpenStudy (mimi_x3):

do you have access to a calculator? or do you want to solve it by hand?

OpenStudy (anonymous):

we are Indians and here most of the time its by hand ...so please try to solve and help me out without the use of calculator,,

hartnn (hartnn):

ok, now can u write 2011*2013 as (2012-1)(2012+1) ?? and do u know what (a+b)(a-b) =??

OpenStudy (anonymous):

yeah \[(a+b)(a-b)=a ^{2}-b ^{2}\]

hartnn (hartnn):

good, so what will be (2012-1)(2012+1) =??

OpenStudy (anonymous):

\[2012^{2}-1\]

hartnn (hartnn):

and u had 1+ 2011*2013 so now whats 1+2012^2-1 ??

OpenStudy (anonymous):

heyyyyy thaaaanzzznnzzzz buddy i got it all after dat ...u made it so easy 4 me now!!!...champ!!!..thnak u!

hartnn (hartnn):

welcome :)

OpenStudy (anonymous):

hey @sriramkumar u typing since long tym eh!

OpenStudy (anonymous):

let 2012=a, then the problemn becomes \[\sqrt{1+(a-4)\sqrt{1+(a-3)\sqrt{1+(a-2)\sqrt{1+(a-1)(a+1)}}}}\] then the inner root becomes \[a ^{2}-1 +1=a ^{2}\] \[1+(a-2)\sqrt{a ^{2}}=1+(a-2)a=1+a ^{2}-2a=(a-1)^{2}\] ...... and so on....

OpenStudy (anonymous):

damn gud!!!....@sriramkumar

OpenStudy (anonymous):

n the answer is 2009

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