Question

$\sqrt[]{1+2005 \sqrt[]{1+2004 \sqrt[]{1+2003*2001}}}$

Answer 1

$\sqrt{1 + 2005 \sqrt{1+2004 \sqrt{1 + 2003*2001}}} = \\ = \sqrt{1 + 2005 \sqrt{1+2004 \sqrt{1 + (2002+1)*(2002-1)}}}=\\= \sqrt{1 + 2005 \sqrt{1+2004 \sqrt{1 + 2002^2 - 1 }}} =\\ = \sqrt{1 + 2005 \sqrt{1+2004 * 2002}} =\\= \sqrt{1 + 2005 * 2003} = 2004$

Answer 2

$\sqrt{1+2005\sqrt{1+2004\sqrt{1+2003\cdot2001}}}=\\ =\sqrt{1+2005\sqrt{1+2004\sqrt{1+2\cdot2001+2001\cdot2001}}}=\\ =\sqrt{1+2005\sqrt{1+2004\sqrt{1+2\cdot1\cdot2001+2001^2}}}=\\ =\sqrt{1+2005\sqrt{1+2004\sqrt{(1+2001)^2}}}=\\ =\sqrt{1+2005\sqrt{1+2004\sqrt{2002^2}}}=\sqrt{1+2005\sqrt{1+2004\cdot2002}}=\\ =\sqrt{1+2005\sqrt{1+2\cdot2002+2002\cdot2002}}=\\ =\sqrt{1+2005\sqrt{1+2\cdot1\cdot2002+2002^2}}=\sqrt{1+2005\sqrt{(1+2002)^2}}=\\ =\sqrt{1+2005\sqrt{2003^2}}=\sqrt{1+2005\cdot2003}=$

$=\sqrt{1+2\cdot2003+2003\cdot2003}=\sqrt{1+2\cdot1\cdot2003+2003^2}=\\ =\sqrt{(1+2003)^2}=\sqrt{2004^2}=2004$