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High School Math : Pre-Calculus

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Example Questions

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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Example Question #1 : Finding Partial Sums In A Series

Find the sum of all even integers from 250 to 350 .

Possible Answers:

15,000

15,500

14,500

14,000

15,300

Correct answer:

15,000

Explanation:

The formula for the sum of an arithmetic series is

$S = \frac{n}{2}(t_{1}+t_{n})$ ,

where is the number of terms in the series, $t_{1}$ is the first term, and $t_{n}$ is the last term.

We know that there are terms in the series. The first term is 250 and the last term is 350 . Our formula becomes:

$S = \frac{50}{2}(250+350)$

S = 25(600) = 15,000

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Example Question #2 : Finding Partial Sums In A Series

Find the sum of all even integers from to .

Possible Answers:

870

930

990

950

900

Correct answer:

930

Explanation:

The formula for the sum of an arithmetic series is

$S = \frac{n}{2}(t_{1}+t_{n})$ ,

where is the number of terms in the series, $t_{1}$ is the first term, and $t_{n}$ is the last term.

$S=\frac{30}{2}(2+60)=15(62)=930$

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Example Question #41 : Pre Calculus

Find the sum of the even integers from 250 to 350 .

Possible Answers:

16,000

15,500

15,300

16,300

15,000

Correct answer:

15,300

Explanation:

The sum of even integers represents an arithmetic series.

The formula for the partial sum of an arithmetic series is

$S = \frac{n}{2}(2(a_1)+(n-1)d)$ ,

where a_1 is the first value in the series, is the number of terms, and is the difference between sequential terms.

Plugging in our values, we get:

$S = \frac{51}{2}(2(250)+(51-1)2)$

S = 25.5(500+100)

S=15,300

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Example Question #41 : Pre Calculus

Find the value for $\sum_{i=0}^{\infty} (\frac{1}{2})^i$

Possible Answers:

1.5

1.25

Correct answer:

Explanation:

To best understand, let's write out the series. So

$\sum_{i=0}^{\infty} (\frac{1}{2})^i = 1+1/2+(1/2)^2 + (1/2)^3+...+(1/2)^\infty$

We can see this is an infinite geometric series with each successive term being multiplied by $\frac{1}{2}$ .

A definition you may wish to remember is

$1+r+r^2+r^3+... +r^\infty= 1/(1-r)$ where stands for the common ratio between the numbers, which in this case is $\frac{1}{2}$ or . So we get

1/(1-r) = 1/(1-.5) = 1/.5 = 2

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Example Question #2 : Finding Sums Of Infinite Series

Evaluate:

$1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343} +...$

Possible Answers:

$\frac{4}{3}$

$\frac{9}{8}$

$\frac{6}{5}$

$\frac{7}{6}$

The series does not converge.

Correct answer:

$\frac{7}{6}$

Explanation:

This is a geometric series whose first term is $a_{0} = 1$ and whose common ratio is $r = \frac{1}{7}$ . The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \frac{1}{7} } = \frac{1}{\frac{7}{7}- \frac{1}{7}}= \frac{1}{\frac{6}{7}} =\frac{7}{6}$

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High School Math : Pre-Calculus

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

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

Example Question #1 : Finding Partial Sums In A Series

Example Question #2 : Finding Partial Sums In A Series

Example Question #41 : Pre Calculus

Example Question #41 : Pre Calculus

Example Question #2 : Finding Sums Of Infinite Series

All High School Math Resources

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