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ariesvang
 2 years ago
Best ResponseYou've already chosen the best response.0howd yuu get 2^11  1 = 2047?

simamura
 2 years ago
Best ResponseYou've already chosen the best response.2This is geometric progression with first member a1=1 and q=4, so sum is 1*(14^6)/(14)=1365

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.0In general, \[ \sum \limits_ {i=0} ^n 2^i = 2^{n+1}1\]

simamura
 2 years ago
Best ResponseYou've already chosen the best response.2@FoolForMath this formula doesn't work here because there is no 8 (2^3) in sum

sahil51993
 2 years ago
Best ResponseYou've already chosen the best response.1Further taking simamura's solution 4^61=4^21(4^4+1+4^2) =15(273) divided by 3 ie 1365

ariesvang
 2 years ago
Best ResponseYou've already chosen the best response.0thank you guys for the help i appreciate it

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.0OKay I can see it now, 1 + 4 + 16 + 64 + 256 + 1024. This is a geometric series with first term 1, and common difference 4.

sheg
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large{1^2+2^2+4^2+8^2+16^2+32^2}\] now you can solve

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2there are only 6 numbers...just add them
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