Find the 7th partial sum of summation of 6 open parentheses 3 close parentheses to the I minus 1 power from 1 to infinity

Question

anonymous · Answer

can you use math notation instead of words?
$$\sum_1^\infty$$

anonymous · Answer

@zepdrix

zepdrix · Answer

$$\Large\rm \sum_{i=1}^{\infty}6(3)^{i-1}$$Is this what the problem should look like?

anonymous · Answer

yes it is

zepdrix · Answer

I think the uhhh geometric series usually starts from an index of 0.
So let's make the substitution $\Large\rm n=i-1$.
So where i starts at 1, n will start at 0 ( 1 - 1).

$$\Large\rm \sum_{n=0}^{\infty}6(3)^{n}$$And so we have the form of a geometric series:$$\Large\rm \sum_{n=0}^{\infty}a r^n$$This series converges when $\Large\rm r\lt1$.
Does that hold true for our problem?

zepdrix · Answer

Oh they wanted us to find the .... 7th partial sum..?

zepdrix · Answer

Hmm I don't know what that is actually :(

zepdrix · Answer

Does that just mean get the first 7 terms of the sum?

anonymous · Answer

yes

zepdrix · Answer

I still think we should make this substitution (unless it's too confusing for you),$$\Large\rm \sum_{n=0}^{\infty}6(3)^{n}$$It just makes it easier to count.

Starting from zero these would be our first 7 terms,$$\Large\rm =6(3)^0+6(3)^1+6(3)^2+6(3)^3+6(3)^4+6(3)^5+6(3)^6$$

zepdrix · Answer

And from there, just use a calculator :U
Understand how I plugged the values in for n?

anonymous · Answer

thank you !!

anonymous · Answer

and yes i did

phi · Answer

There is a formula you can use see https://en.wikipedia.org/wiki/Geometric_series#Formula btw, I think you want to sum from n=0 to n=6 to get the 7th partial sum the answer should be $$ 6 \frac{1-3^7}{1-3} $$