Question

question that is really confusing to me....

Answer 1

\[\sum_{n = 2}^{10}25(0.3)^{n+1}\]

Answer 2

OK HOLD UP

Answer 3

The series goes up to ten and starts at 2, the rate of change is 0.3 and the first value in the series is 25. What do you want to know about it exactly?

Answer 4

It's a geometric series.

Answer 5

well i need to find out how to find thee answer.

Answer 6

Do you know how to calculate partial sums of geometric series?

Answer 7

yeah but i cant get thee right answer

Answer 8

Show work

Answer 9

you got to help me

Answer 10

Use the formula Sn=(a1(1-r^n))/(1-r)
http://www.regentsprep.org/Regents/math/algtrig/ATP2/GeoSeq.htm

Answer 11

a1 is the first value in the series, r is rate, and n is 10 or how far the series goes on.

Answer 12

in your case its 10 btw.

Answer 13

the number above the sigma sign is N

Answer 14

Right now, you have an exponent of \(n+1\), but we want it to be of the form \(k-1\).

So we let \(n+1 = k-1\) which means \(n=k-2\).

Answer 15

The limits, which where \[
2\leq n\leq 10
\]Change to \[
2\leq k-2\leq 10 \implies 4\leq k\leq 12
\]

Answer 16

Series changes to: \[
\sum_{k=4}^{12}\color{red}{25}(\color{blue}{0.3})^{k-1}
\]

Answer 17

Now it is a proper geometric series \(\color{red}{a_1}\color{blue}{r}^{k-1}\)

Answer 18

ok
\[25(0.3)^{n + 1} \]
1.) \[25(0.3)^{2 + 1} = .675\]
2.) \[25(0.3)^{3 + 1} = .2025\]
3.)\[25(0.3)^{4 + 1} = 0.06075\]
4.)\[25(0.3)^{5 + 1} = 0.018225\]
5.)\[25(0.3)^{6 + 1} = 0.0054675\]
and then that is the lowest answer that is shown

Answer 19

One thing we can use is the fact that: \[
\sum_{k=4}^{12}\color{red}{25}(\color{blue}{0.3})^{k-1} =\left( \sum_{k=1}^{12}\color{red}{25}(\color{blue}{0.3})^{k-1} \right)-\left(\sum_{k=1}^{3}\color{red}{25}(\color{blue}{0.3})^{k-1}\right)
\]

Answer 20

\[\sum_{n-2}^{10}25(0.3)^{n+1}\]

Answer 21

Look at the work I did earlier.

Answer 22

oh ok you have to change that for every question to the proper geometric series??

Answer 23

The exponent has to be \(n-1\) for it to be a proper geometric series.

Answer 24

oh i see

Answer 25

There is an alternative method of doing it.

Answer 26

You can also do this:\[
\sum_{n=2}^{10}25(0.3)^{n+1} = \sum_{n=2}^{10}25(0.3)(0.3)^{n} = \sum_{n=2}^{10}25(0.3)^2(0.3)^{n-1}
\]

Answer 27

oh jesus...

Answer 28

But then you have:\[
\sum_{n=2}^{10}\color{red}{25(0.3)^2}(\color{blue}{0.3})^{n-1}
\]

Answer 29

well let me try and can you check if i got the right answer??

Answer 30

Sure.

Answer 31

ok got it is it .964 (rounded to the nearest)

Answer 32

ten thousand

Answer 33

http://www.wolframalpha.com/input/?i=%5Csum_%7Bn+%3D+2%7D%5E%7B10%7D25(0.3)%5E%7Bn%2B1%7D

Looks correct.

Answer 34

thanks