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Precalculus : Pre-Calculus

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Example Questions

Example Question #2 : Sigma Notation

Evaluate: $\sum_{n=6}^{7} n+1$

Possible Answers:

Correct answer:

Explanation:

The summation starts at 6 and ends at 7. Increase the value of after each iteration:

$\sum_{n=6}^{7} n+1 = (6+1)+(7+1) =7+8 = 15$

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Example Question #2 : Sigma Notation

Evaluate: $\sum_{n=1000}^{1001} 2-n$

Possible Answers:

Correct answer:

Explanation:

Rewrite the summation term by term and evaluate.

$\sum_{n=1000}^{1001} 2-n = (2-1000)+(2-1001)$

= -998 -999 = -1997

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Example Question #6 : Sigma Notation

Evaluate: $\sum_{n=2}^{4} -\frac{1}{2n}$

Possible Answers:

$-\frac{1}{4}$

$-\frac{1}{9}$

$-\frac{1}{2}$

$-\frac{13}{24}$

$-\frac{1}{24}$

Correct answer:

$-\frac{13}{24}$

Explanation:

Rewrite the summation term by term:

$\sum_{n=2}^{4} -\frac{1}{2n} = \left(-\frac{1}{2\cdot 2}\right)+ \left(-\frac{1}{2\cdot 3}\right)+\left(-\frac{1}{2\cdot 4}\right)$

$=-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}$

To simplfy we get a common denominator of 24.

$=-\frac{1}{4}\cdot \frac{6}{6}-\frac{1}{6}\cdot \frac{4}{4}-\frac{1}{8}\cdot \frac{3}{3}$

$=-\frac{6}{24}-\frac{4}{24}-\frac{3}{24}=-\frac{13}{24}$

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Example Question #5 : Sigma Notation

Evaluate:

$\sum_{w=3}^{6}(w-1)$

Possible Answers:

Correct answer:

Explanation:

$\sum_{w=3}^{6}(w-1)$ means add the values for w-1 starting with w=3 for every integer until w=6 .

This will look like:

(3-1)+(4-1)+(5-1)+(6-1) = 2 + 3 + 4 + 5 = 14

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Example Question #6 : Sigma Notation

$-2\left[\sum_{n=1}^{5}(n-2)^2\right]$

Possible Answers:

Correct answer:

Explanation:

First, evaluate the sum. We can multiply by -2 last.

The sum

$\sum_{n=1}^{5}(n-2)^2$ means to add together every value for (n-2)^2 for an integer value of n from 1 to 5:

(1-2)^2 + (2-2)^2 + (3-2)^2 + (4-2)^2 + (5-2)^2

=(-1)^2 + 0^2 + 1^2 + 2^2 + 3^2 = 1 + 0 + 1 + 4 + 9 = 15

Now our final step is to multiply by -2.

-2*15 = -30

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Example Question #2 : Sigma Notation

Rewrite this sum using summation notation:

$1+2+4+...+2^{28}$

Possible Answers:

$\sum_{k=0}^{29} 2^{k-1}$

$\sum_{k=1}^{29} 1+2^{k}$

$\sum_{k=1}^{29} 2^{k-1}$

$\sum_{k=1}^{29} 1-2^{k}$

$\sum_{k=1}^{28} 2^{k}$

Correct answer:

$\sum_{k=1}^{29} 2^{k-1}$

Explanation:

First, we must identify a pattern in this sum. Note that the sum can be rewritten as:

$2^0+2^1+2^2+...+2^{28}$

If we want to start our sum at k=1, then the function must be:

$2^{k-1}$ so that the first value is 2^0 .

In order to finish at $2^{28}$ , the last k value must be 29 because 29-1=28.

Thus, our summation notation is as follows:

$\sum_{k=1}^{29} 2^{k-1}$

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Example Question #3 : Sigma Notation

Solve: $\sum_{2}^{4} n+1$

Possible Answers:

Correct answer:

Explanation:

The summation starts at 2 and ends at 4. Write out the terms and solve.

$\sum_{2}^{4} n+1 = (2+1)+ (3+1)+ (4+1)= 3+4+5 =12$

The answer is:

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Example Question #21 : Sequences And Series

Write the following series in sigma notation.

4-9+14-19+...

Possible Answers:

$5\sum_{0}^{n}(-1)^{n}(4)$

$\sum_{0}^{n}(-1)^n(4+5n)$

$\sum_{0}^{n}(-1)^n(4-5n)$

$\sum_{0}^{n}(-1)^{n+1}(4+5n)$

Correct answer:

$\sum_{0}^{n}(-1)^n(4+5n)$

Explanation:

To write in sigma notation, let's make sure we have an alternating sign expression given by:

$\sum_{0}^{n}(-1)^n$

Now that we have the alternating sign, let's establish a function that increases by per term starting at . This is given by

4+5n

Putting it all together,

$\sum_{0}^{n}(-1)^n(4+5n)$

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Example Question #22 : Sequences And Series

Compute: $\sum_{2}^{5} -n+1$

Possible Answers:

Correct answer:

Explanation:

In order to solve this summation, substitute the bottom value of to the function, plus every integer until the iteration reaches to 5.

$\sum_{2}^{5} -n+1 = (-2+1 )+(-3+1) +(-4+1 )+(-5+1)$

(-1)+(-2)+(-3)+(-4) = -10

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Example Question #23 : Sequences And Series

Evaluate: $\sum_{0}^{2} 4\pi+n$

Possible Answers:

$4\pi+3$

$12\pi+3$

$12\pi+2$

$4\pi +2$

$12\pi$

Correct answer:

$12\pi+3$

Explanation:

To evaluate this, input the bottom integer into the expression $4\pi+n$ . Repeat for every integer following the bottom integer until we reach to the top integer . Sum each iteration.

$n=0 :4\pi$

$n=1: 4\pi+1$

$n=2: 4\pi+2$

Add these terms for the summation.

$\sum_{0}^{2} 4\pi+n = 4\pi + (4\pi+1)+ (4\pi +2) = 12\pi+3$

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Sigma Notation

Example Question #2 : Sigma Notation

Example Question #6 : Sigma Notation

Example Question #5 : Sigma Notation

Example Question #6 : Sigma Notation

Example Question #2 : Sigma Notation

Example Question #3 : Sigma Notation

Example Question #21 : Sequences And Series

Example Question #22 : Sequences And Series

Example Question #23 : Sequences And Series

All Precalculus Resources

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