anonymous 4 years ago Evaluate the series 1 + 4 + 16 + 64 + 256 + 1024.

1. anonymous

howd yuu get 2^11 - 1 = 2047?

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

thank you guys for the help i appreciate it

7. anonymous

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.

8. anonymous

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

9. Zarkon

there are only 6 numbers...just add them