Question

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.

Answer 1

@jim_thompson5910 if you're available!

Answer 2

\[\sum_{k=0}^{10} 3 \left(\frac{1}{2} \right)^k\] 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:
```
\[\sum_{k=0}^{10} 3 \left(\frac{1}{2} \right)^k\]
```

Answer 3

yeah that's it :)

Answer 4

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

Answer 5

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

Answer 6

Ok yeah that formula works, do you know what "a" and "r" and "n" are? Can you fill in the question marks?

\[s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{?} ?\]

```
\[s_n=a \frac{1-r^n}{1-r}=\sum_{k=0}^{?} ?\]
```

Answer 7

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

Answer 8

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

Answer 9

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.

\[a\frac{1-r^n}{1-r}=\sum_{k=0}^{n-1} a*r^k\]

Answer 10

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

Answer 11

oh hrm.

Answer 12

so what do we do for the sum formula?