Is there a formula to solve this by hand, or do you actually

>>7727408
derive it. it's not hard.

>>7727408
(1-(8/9)^16)/(1-8/9)

Look up sums of geometric series. Really not that hard.

>>7727427
I'm getting ~7.6 with that. How are they getting 6.6?

>>7727431
I did, but none of the formulas that I found matched this form.. unless I'm just blind

>>7727444
oops, that formula is when n starts at 0, so you have to subtract 8/9 from my answer

>>7727453
That's closer, but I'm still not getting what they got.

>>7727477
Why the fuck are you expressing it in a decimal form in the first place
I meant subtract 1 (= (8/9)^0)

>>7727500
Because the question asked me to.

>>7727523
Disgusting

>>7727552
Blame Pearson and their shitty overpriced excuse for software.

[math]A=\Sigma_{n=1}^m a^n [/math]
[math] a \cdot A=\Sigma_{n=1}^m a^ {n+1} [/math]
[math]a \cdot A=\Sigma_{n=2}^{m+1} a^ {n}[/math]
[math](a-1) \cdot A=a^{m+1}-a [/math]
[math]A=\frac{a^{m+1}-a}{a-1}[/math]

>>7727431
If only I could find my high school algebra text...

>>7727649
I don't think they teach this in high school algebra anymore.

Pre-calculus maybe.

(First term)*(1-reason^(number of terms))/(1-reason)

Test

File: file.png (59KB, 1565x220px) Image search: [Google]

59KB, 1565x220px

(a(1-r^n))/1-r

a is the very first element of the geometric sequence, so for this it would be 8/9.

R is the ratio, for this geometric series the ratio is 8/9.

n=15 for this case etc.

Also, you could have just googled this shit.

>>7727755
Didn't see geometric series until Calc. in college.

>>7727408
It's (8/9)^(15th triangle number)

You're are being asked to find the sum of the first 15 terms. the only way would to do this without a calculator would be if you were given 2 terms in order to be able to distinguish whether it's a geometric or arithmetic sequence then go on from there.

Sum from m to n =( (r)^n+1 -r^n)/r-1

So ((8/9)^n+1 - (8/9)^1)/((8/9)-1)