1. ## Sigma notation help

write the sigma notation of (16/81) , (8/27) , (4/9) , ....

2. Note that $16/2=8$, $8/2=4$, ..., and $81/3=27$, $27/3=9$, ..., (the terms are in a geometric progression) so the $n$th term may be written $a_n=\frac{16}{2^{n-1}}\frac{3^{n-1}}{81}=\frac{16}{81}\left(\frac32\right)^{n-1}$.

Edit: im sorry, that would make your sum $\sum_{n=0}^\infty\frac{16}{81}\left(\frac32\right) ^n$. i hope that helps.

3. i was lookin through that and that seem correct

thanks a bunch

4. Hello, lucifer_x!

Write the sigma notation of (16/81) , (8/27) , (4/9) , ....

"Sigma notation" means that the terms are to be added.

So we have: . $\frac{2^4}{3^4} + \frac{2^3}{3^3} + \frac{2^2}{3^2} + \hdots \;=\;\left(\frac{2}{3}\right)^4 + \left(\frac{2}{3}\right)^3 + \left(\frac{2}{3}\right)^2 + \hdots$

Therefore: . $\sum^{\infty}_{n=1} \left(\frac{2}{3}\right)^{5-n}$

5. oh ya my mind totally blanked cuz i have to turn this in so i was kinda rushing thorugh and scaning through it

u made it so easy to understand , thanks Soroban

when the test comes around i will know what to do