1. anonymous

2. anonymous

sure i just finished a test on it xD

3. anonymous

wit u need help bout?

4. SolomonZelman

It is a geometric sequence, I hope that is not necessary to be mentioned and is well known to all..... right?

5. anonymous

lol if u take the class its a basic

6. SolomonZelman

$$\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}$$ (just putting it for vision, not to look at that tab again...

7. SolomonZelman

$$\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}$$ you need to know: 1) the common ratio. 2) the number of terms that you are adding 3) the (value of the) first term that you start from

8. SolomonZelman

No, you are starting from n=2, Icedragon50

9. anonymous

oooohhh

10. SolomonZelman

The total number of terms is 10-2=8. (But this is not including the $$a_2$$ and so, with the $$a_2$$ you got 9 terms. Your first term that you start adding is $$a_2$$ (then $$a_3$$, $$a_4$$, $$a_5$$ and so on... till the $$a_{10}$$). And to find this first term plug in n=2 into $$25(0.3)^{n+1}$$.

11. SolomonZelman

The common ratio is what you multiply by. You can figure the common ratio (r) simply by writing out a couple of terms. $$\large\color{black}{ \displaystyle a_2=25(0.3)^{2+1}=25(0.3)^3}$$ $$\large\color{black}{ \displaystyle a_2=25(0.3)^{2+2}=25(0.3)^4}$$ $$\large\color{black}{ \displaystyle a_2=25(0.3)^{2+3}=25(0.3)^5}$$ $$\large\color{black}{ \displaystyle a_2=25(0.3)^{2+4}=25(0.3)^6}$$ $$\large\color{black}{ \displaystyle a_2=25(0.3)^{2+5}=25(0.3)^7}$$ And so on... can you give me the common ratio ?

12. anonymous

r=0.3

13. anonymous

Isn't it 0.3

14. SolomonZelman

yes, r=0.3

15. SolomonZelman

So, 9 terms, r=0.3 Now we need our first term, and in that case $$a_2$$

16. anonymous

a1=25(0.3)^2+1

17. SolomonZelman

you need $$a_\color{red}{2}$$

18. SolomonZelman

that is what we start from and not ace of 1.

19. SolomonZelman

$$\large\color{black}{ \displaystyle a_2=25(0.3)^{n+2}=25(0.3)^3=25(27/1000)=27/40}$$

20. SolomonZelman

$$\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}=\frac{(27/40)\cdot (1-0.3^9)}{1-0.3}}$$

21. SolomonZelman

same way you can do the first ten terms w/ $$a_1$$, but to then subtract $$a_1$$ from the sum.

22. anonymous

he's probably just gonna look at the answer since he's not responding so bai bai lol

23. anonymous

My computer actually died and I fell asleep, but ok.