Question

Plzzz Help!
Geometric series:
Find a formula for Sn for:
√3 + 3 + 3√3 + 9 + ... to n terms.

Answer 1

\[S _{n}=\frac{ a(1-r ^{n}) }{ 1-r }\]

Answer 2

a=is the first term in the sequence
r=common ratio

Answer 3

if your common ratio is greater than 1,
\[S _{n}=\frac{ a(r ^{n}-1) }{ r-1 }\]

Answer 4

\[S_{n}=\frac{ \sqrt{3} (\sqrt{3}^{n}-1)}{ \sqrt{3} -1}\]
How do I simplify this?
The answer is not this.

Answer 5

what is the answer?

Answer 6

you have the nth term as
\(a_n=3^{n/2}\)

Answer 7

and are you sure the sum is not to infinity?

Answer 8

the answer is \[S_{n}=\frac{ 3+\sqrt{3} }{ 2 } ((\sqrt{3})^{n} - 1)\]

Answer 9

they rationalized the denominator.
\[\]

Answer 10

\[\frac{\sqrt{3}}{\sqrt{3}-1}\times{\sqrt{3}+1\over\sqrt{3}+1}=\]
simplify

Answer 11

ok thnx a lot @kausarsalley and @electrokid !