\[\sum_{K=1}^{\infty} 4(\frac{ 1 }{ 2 })^{k-1}\]
Evaluate the following
This is a geometric series, right?
Take 4 outside. It becomes an infinite GP with common ratio 1/2
what is the common ratio?
what is the first term?, what do we do when we know these two things?
First term is 1
wait i thought first term was 4?
yes, it is.
Common Ratio 1/2... Calculate sum And multiply the sum by 4 ! 4 is isolated.
4(1/2)^0 4(1) =4
factor out the 4
\[\large \sum_{n=1}^{\infty}a_{1}r^{n-1}\] so a1 =4.
answer would be 8 !! (Final Answer)
The answer is supposed to be 8 but i dont quite get how you get taht
Factor out 4.. and forget it Calculate Sum 1 + 1/2 + 1/4 + 1/8 ...... upto infinity This sum is 2.. Multiply by 4 and you get 8
wait...by "evaluating" are you looking to see if it's convergent/divergent, and if it's convergent,finding the sum?
factor out the 4, then treat it as (1/2)^n and you have first term is 1 then sum is 1/(1-1/2) = 2 then multiply by 4 and get 8
It's obviously convergent..
pretty sure its convergent
it is, because (1/2) <1
so i use the sum of geometric formula then i multiply by 4?
yes
YEs ! @Yellowpanda
okay thank you!
welcome :)
treat it like \[\sum{}{}(\frac{1}{2})^{n}\]
then multiply by 4 because of ab+ac+ad....=a(b+c+d....
Ahh, i see. a1=4, r = 1/2 \[\large S_{n}= \frac{a}{1-r} = \frac{4}{1-(1/2)} = 8\]
And this only applies because the series is convergent....
yes, and note, you could have just factored the 4 out, treated the first term as 1, then multiply the 4 back when you are done. That is what we were saying.
some people get confused on why the ratio is (1/2) when there is a 4 next to it, so I factor it out to show it does not matter.
Ohh I see, I see.
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