Question

Find an explicit rule for the nth term of the sequence.

2, -8, 32, -128, ...

Answer 1

so, what do you think is the common ratio?

Answer 2

*-4

Answer 3

hmm, well, I guess that'd be the actual query.. so
in this geometric sequence is just a multiplier btw
\(\begin{array}{ccccc}
1^{st}&2^{nd}&3^{rd}&4^{th}\\
\hline\\
2,& -8,& 32,& -128,& ....\\
\hline\\
4^{1-1}&4^{2-1}&4^{3-1}&4^{4-1}
\end{array}\)

Answer 4

woops, should be ... -4 rather :(

Answer 5

\( \begin{array}{ccccc}
1^{st}&2^{nd}&3^{rd}&4^{th}\\
\hline\\
2,& -8,& 32,& -128,& ....\\
\hline\\
-4^{1-1}&-4^{2-1}&-4^{3-1}&4^{-4-1}
\end{array}\)

so as you can see, the rule will be \(\bf \large -4^{n-1}\) , n= term ordinal position

Answer 6

shoot I stil have a typo... ohh, well anyhow hehe

Answer 7

so it would be 2*-4^n-1

Answer 8

lol that's okay xD I am understanding

Answer 9

hmm .... yeah... I sorta skipped the a1

Answer 10

thanks :)

Answer 11

yw