What is the sum of a 11–term geometric series if the first term is 4 and the last term is 4,096?

Question

amistre64 · Answer

$$a_n=4\frac{1-r^n}{1-r}$$
i believe is the general setup for this

amistre64 · Answer

or that might be the sum set up :/

callisto · Answer

$$r=\sqrt[10]{4096/4} = 2$$
use the formula above and solve it :)

amistre64 · Answer

$$a_n=4*r^{n-1};\ n=1,2,3,...$$

callisto · Answer

wait, it should be (r^n -1)/(r-1)

amistre64 · Answer

same, you only take a negative out of the top and tranfer it to the bottom or visversa

callisto · Answer

not really lol take the limit and you'll see (i made the same mistake before lol)

anonymous · Answer

ok so what goes where. thats my problemm

amistre64 · Answer

$$\frac{1-r^n}{1-r}*\frac{-1}{-1}=\frac{r^n-1}{r-1}$$

anonymous · Answer

i think you need to find r first maybe?

callisto · Answer

sum = a (r^n -1) / (r-1) 
where r = 2

callisto · Answer

a=4

anonymous · Answer

$$4, 4r, 4r^2, ..., 4r^{10}$$ so you have 
$$4r^{10}=4,096$$
$$r^{10}=1024$$
$$r=\sqrt[10]{1024}=2$$

anonymous · Answer

so r is 2?

anonymous · Answer

oops callisto wrote that already
yes r = 2

amistre64 · Answer

i like how they ask: is that the results?  after youve written out how you get the results ... typoes persist tho

callisto · Answer

alright answer = 8188

anonymous · Answer

$$4\times \frac{2^{11}-1}{2-1}=4(2^{11}-1)$$ is your sum

anonymous · Answer

what callisto said

callisto · Answer

amistre64 when you take the limit it's really different!

anonymous · Answer

ok

anonymous · Answer

8184

callisto · Answer

8188!!!

anonymous · Answer

ok well i was close

amistre64 · Answer

callisto · Answer

callisto · Answer

callisto · Answer

the first post is the mistake i made, the second post is the limit part