What is the slope of the line

Help answering this simple Series Question? I don't know what I did wrong!

Tatiana [17]

Looks like you just evaluated the summand for the given value of $n$ , whereas the question is asking you to find the value of the sum for the first $n$ terms.

Let $S_k=\displaystyle\sum_{n=1}^k\frac3{(-2)^n}$ . Then $S_k$ is the $k$ th partial sum.

$S_1$ happens to be the first term in the series, which is why that box is marked correct:

$S_1=\displaystyle\sum_{n=1}^1\frac3{(-2)^n}=\frac3{(-2)^1}=-1.5$

But the next partial sum is not correct:

$S_2=\displaystyle\sum_{n=1}^2\frac3{(-2)^n}=\frac3{(-2)^1}+\frac3{(-2)^2}=-0.75$

and this is not the same notion as the second term (which indeed is 0.75) in the series.

5 0

3 years ago

daniel is using a scale factor of 10 to enlarge a class photo that measures 3.5 inches by 5 inches what are The dimensions Love

Sholpan [36]

35 inches by 50 inches

6 0

2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$