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Precalculus : Pre-Calculus

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Example Questions

Example Question #1 : Partial Sums Of Series

In case you are not familiar with summation notation, note that: $\sum_{i=1}^{3}(\frac{1}{2})^{i}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$

What is the value of $\sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}$ ?

Possible Answers:

$\frac{5}{8}$

$\frac{15}{8}$

$\frac{31}{16}$

0.825

Correct answer:

$\frac{15}{8}$

Explanation:

Because the iterator starts at , we first have a $\frac{1}{2}^0 = 1$ .

Now expanding the summation to show the step by step process involved in answering the question we get,

$\\ \sum_{i=0}^{3}\left(\frac{1}{2}\right)^{i}=\left(\frac{1}{2}\right)^0+\left(\frac{1}{2}\right)^1+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\\ \\=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\\ \\=\frac{8}{8}+\frac{4}{8}+\frac{2}{8}+\frac{1}{8}\\ \\=\frac{15}{8}$

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Example Question #1 : Finding Sums Of Infinite Series

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

Possible Answers:

1.25

1.5

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 : Sums Of Infinite Series

Evaluate:

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

Possible Answers:

$\frac{9}{8}$

$\frac{4}{3}$

$\frac{7}{6}$

The series does not converge.

$\frac{6}{5}$

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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Example Question #3 : Sums Of Infinite Series

Evaluate:

$1 - \frac{1}{6} + \frac{1}{36} - \frac{1}{216} +...$

Possible Answers:

$\frac{2}{3}$

$\frac{7}{8}$

$\frac{4}{5}$

The series does not converge.

$\frac{6}{7}$

Correct answer:

$\frac{6}{7}$

Explanation:

This is a geometric series whose first term is $a_{0} = 1$ and whose common ratio is $r = - \frac{1}{6}$ . The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \left (- \frac{1}{6} \right )} = \frac{1}{\frac{6}{6}+ \frac{1}{6}}= \frac{1}{\frac{7}{6}} =\frac{6}{7}$

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Example Question #1 : Sums Of Infinite Series

What is the sum of the following infinite series?

$S=2-\frac{1}{3}+1-\frac{1}{9} +\frac{1}{2}-\frac{1}{27}+\frac{1}{4}-\frac{1}{81}+\cdots$

Possible Answers:

diverges

$3\tfrac{1}{2}$

$3\tfrac{5}{9}$

$4\tfrac{1}{2}$

Correct answer:

$3\tfrac{1}{2}$

Explanation:

This series is not alternating - it is the mixture of two geometric series.

The first series has the positive terms.

$2+1+\frac{1}{2}+ \frac{1}{4} + \cdots = 2*\sum\limits_{i=0}^\infty (\frac{1}{2})^i = 2*(\frac{1}{1-\tfrac{1}{2}})=4$

The second series has the negative terms.

$-\frac{1}{3}- \frac{1}{9} -\frac{1}{27}- \cdots = -\frac{1}{3}*\sum\limits_{i=0}^\infty (\frac{1}{3})^i = -\frac{1}{3}*(\frac{1}{1-\tfrac{1}{3}})=-0.5$

The sum of these values is 3.5.

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Example Question #32 : Sequences And Series

What is the sum of the alternating series below?

$10 - 5 + \frac{5}{2} - \frac{5}{4}+\frac{5}{8}-\frac{5}{16}+\cdots$

Possible Answers:

$\frac{20}{3}$

7.5

$\frac{10}{3}$

Correct answer:

$\frac{20}{3}$

Explanation:

The alternating series follows a geometric pattern.

$10 - 5 + \frac{5}{2} - \frac{5}{4} + \cdots = 10*\sum\limits_{i=0}^\infty(-\frac{1}{2})^i$

We can evaluate the geometric series from the formula.

$10*\sum\limits_{i=0}^\infty(-\frac{1}{2})^i = 10*\frac{1}{1+\tfrac{1}{2}} = 10*\frac{2}{3} =\frac{20}{3}$

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Example Question #1 : Sums Of Infinite Series

Find the sum of the following infinite series:

$9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + .....$

Possible Answers:

16.5

13.5

20.25

Correct answer:

13.5

Explanation:

Notice that this is an infinite geometric series, with ratio of terms = 1/3. Hence it can be rewritten as:

$S = \sum_{n=0}^{\infty } 9\cdot \bigg(\frac{1}{3}\bigg)^{n}$

Since the ratio, 1/3, has absolute value less than 1, we can find the sum using this formula:

$S = \frac{a_{0}}{1 - r}$

Where a_0 is the first term of the sequence. In this case $a_0 = 9, r = \frac{1}{3}$ , and thus:

$\frac{9}{1-\frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{9}{1}\cdot \frac{3}{2}=\frac{27}{2} = 13.5$

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Example Question #2 : Sums Of Infinite Series

In the infinite series $a_{1}, a_{2}...a_{n},$ each term $a_{n}=(-2)^{n}$ such that the first two terms are and . What is the sum of the first eight terms in the series?

Possible Answers:

128

-256

-64

210

170

Correct answer:

170

Explanation:

Once you're identified the pattern in the series, you might see a quick way to perform the summation. Since the base of the exponent for each term is negative, the result will be positive if is even, and negative if it is odd. And the series will just list the first 8 powers of 2, with that positive/negative rule attached. So you have:

-2, 4, -8, 16, -32, 64, -128, 256

Note that each "pair" of adjacent numbers has one negative and one positive. for the first pair, -2 + 4 = 2. For the second, -8 + 16 = 8. For the third, -32 + 64 = 32. And so for the fourth, -128 + 256 = 128. You can then quickly sum the values to see that the answer is 170.

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Example Question #1 : Limits

At what value of is the function discontinuous,

$f(x)=\frac{x^2+4}{x-2}$ ?

Possible Answers:

There is no hole

x=2

x=0

x=-2

Correct answer:

x=2

Explanation:

To find discontinuity we need to look at where the denominator of the function is equal to zero. Looking at our function,

$f(x)=\frac{x^2+4}{x-2}$

we need to set the denominator equal to zero and solve for :

x-2=0

x=2

When x = 2 , $f(x)=\frac{8}{0}$ which is undefined.

Therefore x=2 is where the function is discontinuous.

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Example Question #1 : Introductory Calculus

What is the

$\lim_{x \rightarrow 0^-}f(x)=\frac{1}{x}$ ?

Possible Answers:

$\infty$

$-\infty$

The limit does not exist

Correct answer:

$-\infty$

Explanation:

$\lim_{x \rightarrow 0^-}f(x)=\frac{1}{x}$

Means to find the limit of the function as approaches from the left. We can see that x=0 is a vertical asymptote therefore we need to look at numbers extremely close to zero on the left side. The value of continues to decrease to the left of zero, with the value of decreasing even more as gets closer to . Thus, the limit is $-\infty$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Partial Sums Of Series

Example Question #1 : Finding Sums Of Infinite Series

Example Question #2 : Sums Of Infinite Series

Example Question #3 : Sums Of Infinite Series

Example Question #1 : Sums Of Infinite Series

Example Question #32 : Sequences And Series

Example Question #1 : Sums Of Infinite Series

Example Question #2 : Sums Of Infinite Series

Example Question #1 : Limits

Example Question #1 : Introductory Calculus

All Precalculus Resources

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