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Hmmmm.....

Think.. Think.. Think

 

EVALUATE THE EXPRESSION:

 

               \((2+1)(2^2+1)(2^4+1)...(2^{1024}+1)+1\)

 

This tricked me for a while. cheeky

 

 

GOOD LUCK AND MAY THE MATHS BE EVER IN YOUR FAVOR. laugh

 Aug 2, 2016
 #1
avatar+26367 
+5

EVALUATE THE EXPRESSION:
\((2+1)(2^2+1)(2^4+1)...(2^{1024}+1)+1\)

 

\(\begin{array}{|rcll|} \hline (2^{\color{red}1}+1)(2^{\color{red}2}+1) &=& 2^{\color{red}3}+2^2+2^1+1 &| 1+2 = 3\\ (2^{\color{red}1}+1)(2^{\color{red}2}+1)(2^{\color{red}4}+1)&=& 2^{\color{red}7}+2^6+2^5+2^4+2^3+2^2+2^1+1 &| 1+2+4=7\\ \dots \\ (2^1+1)(2^2+1)(2^4+1)...(2^{1024}+1) &=& 2^{2047}+2^{2046}+\dots +2^1 + 1 \\ &&|1+2+4+\dots + 1024 = 2047 \\\\ & & 2^{2047}+2^{2046}+\dots +2^1 + 1 = 2^{2048}-1 \\\\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ (2^1+1)(2^2+1)(2^4+1)...(2^{1024}+1)+1 } & \mathbf{=} & \mathbf{2^{2048}}\\ \hline \end{array}\)

 

laugh

 Aug 2, 2016
 #2
avatar
0

CPhill: Can you please explain this result using Wolfram/Alpha? Where am I going wrong?

∏ [ 2^(2n) + 1], n=0 to 1023

=product_(n=0)^1023(1+2^(2 n))≈1.016900708110399312582356305677699335401×10^315,345

 Aug 2, 2016

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