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High School Math : High School Math

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Example Questions

Example Question #2 : Using Sigma Notation

Indicate the sum of the following series:

$\sum_{m=-2}^{9}(2)^m$

Possible Answers:

1230.75

1302.75

1000.25

1023.75

1032.25

Correct answer:

1023.75

Explanation:

The formula for the sum of a geometric series is

$S=\frac{a_1-a_1r^n}{1-r}$ ,

where a_1 is the first term in the series, is the rate of change between sequential terms, and is the number of terms in the series

For this problem, these values are:

$a_1 = \frac{1}{4}$

r = 2

n = 12

Plugging in our values, we get:

$S=\frac{\frac{1}{4}-\frac{1}{4}(2)^{12}}{1-2}$

S = 1023.75

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Example Question #1 : Using Sigma Notation

Indicate the sum of the following series.

$\sum_{b=2}^{11} \frac{1}{2}(4)^{b-2}$

Possible Answers:

172,764.5

174,726.5

176,742.5

164,772.5

174,762.5

Correct answer:

174,762.5

Explanation:

The formula for the sum of a geometric series is

$S=\frac{a_1-a_1r^n}{1-r}$ ,

where a_1 is the first term in the series, is the rate of change between sequential terms, and is the number of terms in the series

In this problem we have:

$a_1 = \frac{1}{2}$

r = 4

n = 10

Plugging in our values, we get:

$S=\frac{\frac{1}{2}-\frac{1}{2}(4)^{10}}{1-4}$

S = 174,762.5

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Example Question #181 : Pre Calculus

Consider the sequence: $2,\ 5,\ 8,\ 11,\ ...$

What is the fifteenth term in the sequence?

Possible Answers:

Correct answer:

Explanation:

The sequence can be described by the equation $\small 2+3(n-1)$ , where is the term in the sequence.

For the 15th term, n=15 .

$\small 2+3(n-1)=2+3(15-1)$

2+3(14)

2+42

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Example Question #12 : Sequences And Series

What are the first three terms in the series?

$\sum_{n=1}^{7}(4n-5)$

Possible Answers:

(1)+(3)+(7)

(-5)+(-1)+(3)

(-1)+(3)+(8)

(-1)+(3)+(7)

(3)+(7)+(11)

Correct answer:

(-1)+(3)+(7)

Explanation:

To find the first three terms, replace with , , and .

(4n-5) = (4(1)-5)=-1

(4n-5) = (4(2)-5)=3

(4n-5) = (4(3)-5)=7

The first three terms are , , and .

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Example Question #1 : Finding Terms In A Series

Find the first three terms in the series.

$\sum_{n=3}^{10}(8-5n)$

Possible Answers:

(3)+(-2)+(-7)

(-2)+(-7)+(-12)

(-7)+(12)+(-17)

(-7)+(-12)+(-17)

(7)+(12)+(17)

Correct answer:

(-7)+(-12)+(-17)

Explanation:

To find the first three terms, replace with , , and .

(8-5n) = (8-5(3))=-7

(8-5n) = (8-5(4))=-12

(8-5n) = (8-5(5))=-17

The first three terms are , , and .

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Example Question #41 : Pre Calculus

Indicate the first three terms of the following series:

$\sum_{x=1}^{7}(4x-5)$

Possible Answers:

(1)+(-3)+(-7)

(-1)+(3)+(7)

(1)+(-3)+(1)

(-1)+(3)+(-1)

(1)+(5)+(9)

Correct answer:

(-1)+(3)+(7)

Explanation:

In the arithmetic series, the first terms can be found by plugging , , and into the equation.

x=1

4(1)-5 = -1

x=2

4(2)-5 = 3

x=3

4(3)-5 = 7

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Example Question #2051 : High School Math

Indicate the first three terms of the following series:

$\sum_{c=3}^{10}(8-5c)$

Possible Answers:

(7)+(-12)+(-17)

(7)+(-12)+(17)

(-7)+(-12)+(-17)

(-7)+(12)+(-17)

(7)+(12)+(17)

Correct answer:

(-7)+(-12)+(-17)

Explanation:

In the arithmetic series, the first terms can be found by plugging in , , and for .

c=3

8-5(3) = -7

c=4

8-5(4) = -12

c=5

8-5(5) = -17

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Example Question #3 : Finding Terms In A Series

Indicate the first three terms of the following series:

$\sum_{m=-2}^{9}(2)^m$

Possible Answers:

$(-\frac{1}{4})+(-\frac{1}{2})+(1)$

$(\frac{1}{4})+(\frac{1}{2})+(0)$

(4)+(2)+(1)

$(-\frac{1}{4})+(-\frac{1}{2})+(-1)$

$(\frac{1}{4})+(\frac{1}{2})+(1)$

Correct answer:

$(\frac{1}{4})+(\frac{1}{2})+(1)$

Explanation:

The first terms can be found by substituting , , and for into the sum formula.

m=-2

$(2)^{-2}=\frac{1}{4}$

m=-1

$(2)^{-1}=\frac{1}{2}$

m=0

$(2)^{0}=1$

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Example Question #4 : Finding Terms In A Series

Indicate the first three terms of the following series.

$\sum_{b=2}^{11} \frac{1}{2}(4)^{b-2}$

Possible Answers:

$\frac{1}{2}\cdot 2\cdot 8$

Not enough information

$\frac{1}{4}\cdot 4\cdot 16$

$\frac{1}{2}+2+8$

$\frac{1}{4}+4+16$

Correct answer:

$\frac{1}{2}+2+8$

Explanation:

The first terms can be found by substituting , , and in for .

b=2

$\frac{1}{2}(4)^{2-2}=\frac{1}{2}$

b=3

$\frac{1}{2}(4)^{3-2}=2$

b=4

$\frac{1}{2}(4)^{4-2}=8$

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Example Question #5 : Finding Terms In A Series

What is the sixth term when $\left (\frac{1}{2}x-2 \right )^{10}$ is expanded?

Possible Answers:

252x^5

-840x^5

840x^6

-252x^5

-840x^6

Correct answer:

-252x^5

Explanation:

We will need to use the Binomial Theorem in order to solve this problem. Consider the expansion of (a+b)^n , where n is an integer. The rth term of this expansion is given by the following formula:

$\binom{n}{r-1}a^{n-r+1}b^{r-1}$ ,

where $\binom{n}{r-1}$ is a combination. In general, if x and y are nonnegative integers such that x > y, then the combination of x and y is defined as follows: $\binom{x}{y}=_xC_y=\frac{x!}{(y)!(x-y)!}$ .

We are asked to find the sixth term of $\left (\frac{1}{2}x-2 \right )^{10}$ , which means that in this case r = 6 and n = 10. Also, we will let $a=\frac{1}{2}x$ and b=-2 . We can now apply the Binomial Theorem to determine the sixth term, which is as follows:

$\binom{10}{6-1}\left (\frac{1}{2}x \right )^{10-6+1}\cdot(-2) ^{6-1}$

$=\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$

Next, let's find the value of $\binom{10}{5}$ . According to the definition of a combination,

$\binom{10}{5}=\frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5!}{5!5!}$

$=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=252$ .

Remember that, if n is a positive integer, then $n!=n\cdot(n-1)\cdot(n-2)\cdot\cdot\cdot3\cdot2\cdot1$ . This is called a factorial.

Let's go back to simplifying $\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$ .

$\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}=252\cdot\left (\frac{1}{2}x \right )^{5}\cdot-32$

$=252\cdot\left (\frac{1}{2} \right )^5\cdot x^5\cdot (-32) =252\cdot \frac{1}{32}\cdot-3 2\cdot x^5$

=-252x^5

The answer is -252x^5 .

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #2 : Using Sigma Notation

Example Question #1 : Using Sigma Notation

Example Question #181 : Pre Calculus

Example Question #12 : Sequences And Series

Example Question #1 : Finding Terms In A Series

Example Question #41 : Pre Calculus

Example Question #2051 : High School Math

Example Question #3 : Finding Terms In A Series

Example Question #4 : Finding Terms In A Series

Example Question #5 : Finding Terms In A Series

All High School Math Resources

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