Geometric series formula for 1/3+1/3^2+1/3^3+...+1/3^n

[JCCol]

I am new to number theory and i so far like it. I recently found this problem but there was no explanation of how they got it and i was wondering if any of you knew how they got it. It is this:

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

Then they used:

1/4+1/4^2+1/4^3... + 1/4^n=1/3(1-1/4^n)

Now what i don't see is how they got that final equation (probably a dumb question) but i would appreciate it if anyone could help. Thanks.

 

[LeonhardEuler]

I am new to number theory and i so far like it. I recently found this problem but there was no explanation of how they got it and i was wondering if any of you knew how they got it. It is this:

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

Then they used:

1/4+1/4^2+1/4^3... + 1/4^n=1/3(1-1/4^n)

Now what i don't see is how they got that final equation (probably a dumb question) but i would appreciate it if anyone could help. Thanks.

It's the formula for a geometric series. Suppose we have a series:
$$\sum_{j=0}^{n}ar^j=a+ar+ar^2+ar^3+...+ar^n=a(1+r+r^2+r^3+...+r^n)$$
Just multiply and divide by (1-r) to get:
$$=a\frac{(1+r+r^2+r^3+...+r^n)(1-r)}{1-r}$$
When you expand out the multiplication on the top, most of the terms cancel:
$$=a\frac{1+r+r^2+r^3+...+r^n - r - r^2 - r^3 -r^4 -...-r^{n+1}}{1-r}$$
Canceling:
$$=a\frac{1-r^{n+1}}{1-r}$$
Now the first sum is just the sum
$$\sum_{j=0}^{n}ar^j$$
with a=(1/3) and r= (1/3) while in the second sum a and r are (1/4). Do out the subtraction on the denomenator and reciprocate to turn the division into multiplication and you'll come out with what you gave.

By the way, using math typesetting can make your posts clearer. Click on any equation to see how it was made. It is not too difficult to pick up on how to use.

 

[JCCol]

Thank you this explains a lot and i will try to use math typeset.