Question

Sigma Notation: Need Help!!
Problem 1:
Evaluate the summation of 3 times negative 2 to the n minus 1 power, from n equals 1 to 5.

Choices:
A.-93
B.-33
C.33
D.93

Answer 1

I don't know how to do this and would appreciate help and an explanation.

Answer 2

\[\sum_{n=1}^{5} 3(-2) ^ n-1\]

Answer 3

there may be a faster way but as far as i can see, just plug in the "n" values and add up the results

Answer 4

I don't know how :/

Answer 5

the 3 can go on outside since its just a multiplier
\[\rightarrow 3\sum_{n=1}^{5}(-2)^{n-1}\]

Answer 6

The way I wrote it is how it is set up

Answer 7

well what is (-2)^0?
(-2)^1
and so on

Answer 8

oh sorry so the "-1" is not part of the exponent?

Answer 9

\[
\sum_{n=1}^{5} 3(-2) ^ n - 1 = \\
\sum_{n=1}^{5} 3(-2) ^ n - \sum_{n=1}^{5}1 = \\
3\sum_{n=1}^{5} (-2) ^ n - 5 \\
3(-2 + 4 - 8 + 16 - 32) - 5 = ?
\]

Answer 10

not that I am aware of I wrote it as it is written. I don't know the steps or how to solve it..

Answer 11

hmm that give -71 which is not an answer choice
i think its meant to be as I wrote it

Answer 12

I don't know. its on my work which I have like 4 probs like this involving sigma notation and that's exactly how it is written.

Answer 13

\[
\sum_{n=1}^{5} 3(-2) ^ {n - 1} = 3\sum_{n=1}^{5} (-2) ^{n-1} = \\
3(1 - 2 + 4 - 8 + 16) = 3 = 33
\]

Answer 14

There is also this:
Evaluate the summation of 2 n plus 5, from n equals 1 to 12.
Choices:
A.29
B.36
C.216
D.432
@aum

Answer 15

You have to clarify if 2 is raised to an exponent or 2 is multiplied by n. Is 5 in the exponent?

Answer 16

Use ^ for exponents. Use parethesis to group items if necessary.

Answer 17

I have two more after this one if you could help.
The one I just posted looks like this:
\[\sum_{n=1}^{12} 2n+5\]

Answer 18

\[
\sum_{n=1}^{12}2n+5 = \sum_{n=1}^{12}2n + \sum_{n=1}^{12}5 = \\
2\sum_{n=1}^{12} n + (5 * 12) = 2\frac{12*13}{2} + 60 = ?
\]

Answer 19

How do I do math to solve top one. I know 5*12=60 and 12*13/2 = 78 then when you add 60 it is 138

Answer 20

The 2 on the top and bottom will cancel. Just (12 * 13) + 60

Answer 21

so 216? is that the answer? and can you help and explain with two more?

Answer 22

For the above problem I have used the formula for adding:
1 + 2 + 3 + 4 + ..... + n = n(n+1) / 2

Here n = 12. So 12 * 13 / 2 is the sum

To that sum add 60 to get 216.

Yes, 216.

Answer 23

go ahead.

Answer 24

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

Answer 25

What do you mean n+1 is small.
Are both n and 1 in the exponent?

Answer 26

yes exactly like their the power of or to n+1

Answer 27

Then you should write like this:
\[\large \sum_{n=2}^{10}25(0.3)^{n+1} \]

Answer 28

yes exactly how it looks I just didn't know how to make both small. Only the n is what I knew how to do.

Answer 29

Right click on the formula I wrote and click on "Show Math as" and then "Tex commands". It will show the exact latex code that anyone uses to write their stuff. Just the n+1 should be enclosed in {n+1} to have both as exponents.

Answer 30

oh okay I gotcha so how would we set this up to solve this one. After this one theres one more then you can go, I appreciate the help!!

Answer 31

\[
\large \sum_{n=2}^{10}25(0.3)^{n+1} = 25\sum_{n=2}^{10}(0.3)^{n+1} = \\ \large
25[0.3^3 + 0.3^4 + 0.3^5 + .... +0.3^{11}] = 25 * [~~??~~]
\]
Use the formula for the sum of a geometric sequence to add up \(\large [0.3^3 + 0.3^4 + 0.3^5 + .... +0.3^{11}]\)

Answer 32

-Aum can you please help me with sample size

Answer 33

is it 3.3? @aum

Answer 34

What are your answer choices?

Answer 35

A.0.005
B.0.321
C.0.964
D.10.714

Answer 36

Sample size of +- 7%

Answer 37

I got 3.0375 I don't get it/: what did I do wrong? or what am I not doing?

Answer 38

Use the formula for the sum of a geometric sequence to add up \(\large [0.3^3 + 0.3^4 + 0.3^5 + .... +0.3^{11}]\)

The formula for adding n terms of a geometric sequence,
a + ar + ar^2 + ....+ar^(n-1) is: a * (1 - r^n) / (1 - r).

We need to add 0.3^3 + 0.3^4 + 0.3^5 + ..... + 0.3^(11)

Here the first term a = 0.3^3
common ratio r = 0.3
number of terms when n goes from 2 to 10 is 9.

Plug the numbers into the geometric sum formula.

Then multiply the sum by 25 to get the answer.

Answer 39

Bruh!

Answer 40

sorry im trying to add them up but its not making sense, when I get to multiply 25 it hikes up to 29

Answer 41

\[
\large
25[0.3^3 + 0.3^4 + 0.3^5 + .... +0.3^{11}] = 25 * 0.3^3\frac{1 - 0.3^9}{1-0.3} = \\ \large
25 * 0.027 *\frac{0.99998}{0.7} = ?
\]

Answer 42

The last problem I ll need help with will look like this if you wanna look at it while I figure this one out.

\[\sum_{n=3}^{12} -5n-1\]
And choices are:
A.-410
B.-385
C.410
D.385

Answer 43

is it C.0.964?

Answer 44

\[ \large
\sum_{n=3}^{12} -5n-1 = -5\sum_{n=3}^{12} n -\sum_{n=3}^{12} 1 = \\ \large
-5 * \frac{(3+12)}{2} * 10 - 10 = ?
\]

Answer 45

Yes, C 0.964 for the previous problem.

Answer 46

-385? right?:D

Answer 47

yes.

Answer 48

WOO thank you @aum your a life saver!

Answer 49

You are welcome.