## Babynini one year ago Find the partial sum S_n of the geometric sequence that satisfies the given conditions. 10 over sigma k=0 beneath 3(1/2)^k on the side.

1. Babynini

@jim_thompson5910 if you're available!

2. Empty

$\sum_{k=0}^{10} 3 \left(\frac{1}{2} \right)^k$ Right? Any ideas how to simplify this? Do you know the geometric series formula or know how to derive it? here's the code by the way:  $\sum_{k=0}^{10} 3 \left(\frac{1}{2} \right)^k$ 

3. Babynini

yeah that's it :)

4. Babynini

er they've had me using the formula s_n=a([1-r^n]/[1-r]) for the sum.

5. Babynini

do you mean if I know how to write that out in terms?

6. Empty

Ok yeah that formula works, do you know what "a" and "r" and "n" are? Can you fill in the question marks? $s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{?} ?$  $s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{?} ?$ 

7. Babynini

oh oh I see. $s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{n} ?$ like that?

8. Babynini

$s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{n} a(r)$ is that correct? o.o

9. Empty

Almost not quite, since if you plug in n=1 you would get what? $a \frac{1-r^1}{1-r} = 1$ for the left side and $\sum_{k=0}^{1} a(r) = ar+ar=2ar$ on the right side. I'll give you a hint, try this, add a power of k in there: (for this example I picked 2 on the top to show how I calculate it on the right, but it also has a geometric series that I haven't shown you that it's equal to.) $\sum_{k=0}^{2} a*r^k= ar^0+ar^1+ar^2$ But the real value on top will be n-1, not n. $a\frac{1-r^n}{1-r}=\sum_{k=0}^{n-1} a*r^k$

10. Babynini

ah, sorry. Pc got messed up but i'm back now!

11. Babynini

oh hrm.

12. Babynini

so what do we do for the sum formula?