Akshit_math
The value of _____
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hartnn
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is _______
Akshit_math
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\[\sqrt{1 + 2008\sqrt{1 + 2009\sqrt{1 + 2010\sqrt{1 + 2011.2013}}}}\]
Akshit_math
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what is the value of the above ???
Mimi_x3
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do you have access to a calculator?
or do you want to solve it by hand?
Akshit_math
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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
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ok, now can u write 2011*2013 as (2012-1)(2012+1) ??
and do u know what (a+b)(a-b) =??
Akshit_math
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yeah \[(a+b)(a-b)=a ^{2}-b ^{2}\]
hartnn
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good, so what will be (2012-1)(2012+1) =??
Akshit_math
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\[2012^{2}-1\]
hartnn
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and u had 1+ 2011*2013
so now whats 1+2012^2-1 ??
Akshit_math
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heyyyyy thaaaanzzznnzzzz buddy i got it all after dat ...u made it so easy 4 me now!!!...champ!!!..thnak u!
hartnn
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welcome :)
Akshit_math
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hey @sriramkumar u typing since long tym eh!
sriramkumar
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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....
Akshit_math
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damn gud!!!....@sriramkumar
Akshit_math
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n the answer is 2009