1+3+3^2+....+3^49 is 3^49 -1 or 3^50 -1 ?

Question

mariamikayla · Answer

$$3^{49}-1 \ \ or \ \ 3^{50} -1$$

agent0smith · Answer

1+3+3^2+....+3^49

already includes 3^49, plus 3^48 and a bunch of other terms far greater than 1... so how could it possibly equal 3^49 - 1?

mariamikayla · Answer

Using geometrical progression it gives me http://prntscr.com/bsgm2y it is right?

agent0smith · Answer

I think you might want to post a screenshot of the question.

agent0smith · Answer

1+3+3^2+....+3^49 Is a geometric series with a1 = 1 and r = 3 and n=50. Your original question makes very little sense due to your wording. If you need the sum of 1+3+3^2+....+3^49, then your work http://prntscr.com/bsgm2y looks fine

agent0smith · Answer

http://www.regentsprep.org/regents/math/algtrig/atp2/ArithG12.gif

mariamikayla · Answer

Thanks a lot :D i've got it now ;)

sshayer · Answer

$$1+3+3^2+...+3^{49}=3^0+3^1+3^2+...+3^{49}$$
number of terms=50
$$a=1,r=3,s _{50}=a \left( \frac{ r^n-1 }{ r-1 } \right)=1 \left( \frac{ 3^{50}-1 }{ 3-1 } \right)=?$$