The expression $(2^{2^{2014}}+1)(2^{2^{2013}}+1)\cdots(2^{2^2}+1)(2^{2^1}+1)(2^{2^0}+1)+1$ evaluates to a specific integer value through the application of algebraic identities. The product of terms of the form $(2^{2^n}+1)$ for \( n = 0 \) to \( 2014 \) can be simplified using the formula for the sum of powers of two. This results in a final expression that can be computed without a calculator, demonstrating the power of mathematical reasoning over computational tools.
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anemone said:Evaluate the expression $(2^{2^{2014}}+1)(2^{2^{2013}}+1)\cdots(2^{2^2}+1)(2^{2^1}+1)(2^{2^0}+1)+1$
kaliprasad said:wish you happy new year
multiply 1st term by $(2^{2^0}-1) $ which is nothing but 1
and using
$(2^{2^{n-1}}+1)(2^{2^{n-1}}-1) = (2^{2^{n}}-1)$ from n = 1 to 2015 one by one
to get
$(2^{2^{2015}}-1+1)$ or $2^{2^{2015}}$