Question

Can someone please help me with this Sigma Notation question?????

Answer 1

sure i just finished a test on it xD

Answer 2

wit u need help bout?

Answer 3

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

Answer 4

lol if u take the class its a basic

Answer 5

\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\)

(just putting it for vision, not to look at that tab again...

Answer 6

\(\large\color{black}{ \displaystyle \sum_{ n=2 }^{ 10 } ~25(0.3)^{n+1}}\)

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

Answer 7

No, you are starting from n=2, Icedragon50

Answer 8

oooohhh

Answer 9

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 \(25(0.3)^{n+1}\).

Answer 10

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 ?

Answer 11

r=0.3

Answer 12

Isn't it 0.3

Answer 13

yes, r=0.3

Answer 14

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

Answer 15

a1=25(0.3)^2+1

Answer 16

you need \(a_\color{red}{2}\)

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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