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Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{1 + 2008\sqrt{1 + 2009\sqrt{1 + 2010\sqrt{1 + 2011.2013}}}}\]
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
what is the value of the above ???
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.0
do you have access to a calculator? or do you want to solve it by hand?
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
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,,
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
ok, now can u write 2011*2013 as (20121)(2012+1) ?? and do u know what (a+b)(ab) =??
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
yeah \[(a+b)(ab)=a ^{2}b ^{2}\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
good, so what will be (20121)(2012+1) =??
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
\[2012^{2}1\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
and u had 1+ 2011*2013 so now whats 1+2012^21 ??
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
heyyyyy thaaaanzzznnzzzz buddy i got it all after dat ...u made it so easy 4 me now!!!...champ!!!..thnak u!
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
hey @sriramkumar u typing since long tym eh!
 one year ago

sriramkumar Group TitleBest ResponseYou've already chosen the best response.1
let 2012=a, then the problemn becomes \[\sqrt{1+(a4)\sqrt{1+(a3)\sqrt{1+(a2)\sqrt{1+(a1)(a+1)}}}}\] then the inner root becomes \[a ^{2}1 +1=a ^{2}\] \[1+(a2)\sqrt{a ^{2}}=1+(a2)a=1+a ^{2}2a=(a1)^{2}\] ...... and so on....
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
damn gud!!!....@sriramkumar
 one year ago

Akshit_math Group TitleBest ResponseYou've already chosen the best response.0
n the answer is 2009
 one year ago
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