Find the sum

1/7 + 3/7 + 3^2 / 7 + 3^3/7 + ..... 3^(n-1) 7

Question

Find the sum

1/7 + 3/7 + 3^2 / 7 + 3^3/7 + ..... 3^(n-1) 7

anonymous · Answer

Hey take 1/7 common out of all terms... you will notice a geometric progression formed...

anonymous · Answer

$$\frac{1}{7} (3^0 + 3^1 + 3^2 ..... 3^{(n-1)}) $$

I hope you are familiar with the sum of geometric progression.

$$\frac{a(1-r^n)}{1-r} $$

Take 1 as a ; 3 as r

anonymous · Answer

It says that the answer its $$-\frac{ 1 }{ 14 } (1-3^{n})$$

anonymous · Answer

Did you try solving it?

anonymous · Answer

Yes the answer is 4

anonymous · Answer

Wait! Answer is 4 ? , How did you get it ?

anonymous · Answer

$$\frac{ 1(1-3^{2}) }{ 1-3 } = \frac{ -8 }{ -2 } = 4$$

anonymous · Answer

According to the question, you don't know the value of n , so, you can't just put n as 2 ! :P :) 
Moreover , you missed out that 1/7 :P

anonymous · Answer

Im lost :( sorry. So it will be $$\frac{ 1(1-3^{n}) }{ 1-3 }$$ What I do after

anonymous · Answer

It's okay to be lost! :D

And, what you did, isn't wrong, it's just incomplete. 

it would be : 

$$\frac{1}{7}\frac{1(1−3^n)}{1−r}$$

anonymous · Answer

Now I get it! Your awersome :)

anonymous · Answer

Haha. :D Thank You! :D