Evaluate the series 1 + 4 + 16 + 64 + 256 + 1024.

Question

anonymous · Answer

howd yuu get 2^11 - 1 = 2047?

anonymous · Answer

This is geometric progression with first member a1=1 and q=4, so sum is 1*(1-4^6)/(1-4)=1365

anonymous · Answer

In general, 
$$ \sum \limits_ {i=0} ^n 2^i = 2^{n+1}-1$$

anonymous · Answer

@FoolForMath this formula doesn't work here because there is no 8 (2^3) in sum

anonymous · Answer

Further taking simamura's solution
4^6-1=4^2-1(4^4+1+4^2)
=15(273) divided by 3
ie 1365

anonymous · Answer

thank you guys for the help i appreciate it

anonymous · Answer

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.

anonymous · Answer

$$\large{1^2+2^2+4^2+8^2+16^2+32^2}$$
now you can solve

zarkon · Answer

there are only 6 numbers...just add them