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ariesvang
Evaluate the series 1 + 4 + 16 + 64 + 256 + 1024.
howd yuu get 2^11 - 1 = 2047?
This is geometric progression with first member a1=1 and q=4, so sum is 1*(1-4^6)/(1-4)=1365
In general, \[ \sum \limits_ {i=0} ^n 2^i = 2^{n+1}-1\]
@FoolForMath this formula doesn't work here because there is no 8 (2^3) in sum
Further taking simamura's solution 4^6-1=4^2-1(4^4+1+4^2) =15(273) divided by 3 ie 1365
thank you guys for the help i appreciate it
OKay I can see it now, 1 + 4 + 16 + 64 + 256 + 1024. This is a geometric series with first term 1, and common difference 4.
\[\large{1^2+2^2+4^2+8^2+16^2+32^2}\] now you can solve
there are only 6 numbers...just add them