Sigma Notation - Algebra II

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Question

$\sum_{n=1}^{5}2n-3$

What is the sum of the series?

Answer

$\sum_{n=1}^{5}2n-3$

The number on the bottom is the first value we plug in for n. We keep substituting higher and higher integer values of n until we get to the top number (in this case, 5).

Here's what it looks like:

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Question

$\sum_{n=1}^{\infty }\frac{1}{2^n}$

What is the sum of the series?

Answer

Recall the formula for the sum of an infinite geometric series:

$S = \frac{a}{1-r}$

Of course, this formula only works if the series is geometric with a common ratio between -1 and 1.

While the sigma notation calls for substituting an infinite number of values for n, let's just substitute a few to see if we can find a pattern:

$\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...$

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...$

From this, we can see that we have a geometric series with a common ratio of 1/2 and a first term of 1/2.

$S = \frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

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Question

Solve for if $x = \sum i^{2}$ for to

Answer

For summations, we evaluate the expression at each value of , then add all of the results together.

For this problem, we are working from to .

Then adding everything up, we get

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Question

Evaluate the following summation:

$\sum_{n=1}^{5}\frac{n^2+4}{2n}$

Answer

According to the notation in the problem, we are told to sum the results obtained by evaluating the equation at each integer between the numbers below and above the sigma. For the sigma notation of this problem in particular, this means we start by plugging 1 into our equation, and then add the results obtained from plugging in 2, and then 3, and then 4, stopping after we add the result obatined from plugging 5 into the equation, as this is the number on top of sigma at which we stop the summation. This process is worked out below:

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

$=\frac{5}{2}+2+\frac{13}{6}+\frac{5}{2}+\frac{29}{10}=\frac{75}{30}+\frac{60}{30}+\frac{65}{30}+\frac{75}{30}+\frac{87}{30}$

$=\frac{362}{30}=\frac{181}{15}$

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Question

Evaluate: $\sum_{n=1}^{5} n$

Answer

The summation starts from 1 and ends at 5. This can be rewritten as:

$\sum_{n=1}^{5} n =1+2+3+4+5=15$

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Question

Evaluate: $\sum_{n=1}^{5} n^2$

Answer

The summation starts from 1 and ends at 5. Rewrite the summation sign:

$\sum_{n=1}^{5} n^2=1^2+2^2+3^2+4^2+5^2= 1+4+9+16+25=55$

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Question

Evaluate: $\sum_{n=0}^{3} \ln(n)$

Answer

The natural log domain is only valid for values greater than zero. Therefore, the solution does not exist.

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Question

Evaluate: $\sum_{n=1}^{5} (-1)^n (2^n)$

Answer

Rewrite the summation starting from 1 to 5 and add the terms.

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

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Question

The Fibonacci sequence is given as

.

.

.

Find

$\sum_{i=1}^{4}a_i$ .

Answer

$\sum_{i=1}^{4}a_i$ is defined to be the sum of the terms from $1\leq i\leq 4$ for $i\epsilon \mathbb{Z}$ .

In other words,

$\ \sum_{i=1}^{4}a_i \ \= a_1+a_2+a_3+a_4\ \=1+1+2+3\ \=7$

As such, the asnwer is .

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Question

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

Answer

The sigma notation indicates summation starting from 1 and ends at 4.

Substitute in order of the iteration.

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

$\sum_{1}^{4} n+2 = 18$

The correct answer is .

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Question

Summation (Sigma) Notation

Rewrite the below series in summation or sigma notation:

Answer

First we must write an equation for the pattern in the series. We can see that this is an arithmetic pattern because the same number (4) is being added each time.

The formula to write the rule or equation for an arithmetic pattern is:

$a{_{n}}=a{_{1}}+(n-1)d$

Where a1 is the first value in the sequence and d is the common difference, n is left as a variable.

When we plug in the values we are given we get:

$a{_{n}}=1+(n-1)4$

Which simplifies to become:

$a{_{n}}=1+4n-4$

Which further simplifies to become:

$a{_{n}}=4n-3$

Now we must put this in the sigma or summation notation.

We know that summation notation always involves the greek letter sigma

$\sum$

The number on the bottom is the first number that is plugged in, first input or first domain number - in this case it is 1.

The number on top is the last number that is plugged in, the last input or the number of numbers in the series - there are five numbers in this series, so it ends with 5.

So the final summation will look like this:

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

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Question

Solve: $\sum_{0}^{2} (n-1 )- \sum_{1}^{3} n$

Answer

Evaluate each summation term first.

The bottom number is the first term of the summation. Plug the number into the expression. Repeat the summation for every whole number until the summation reaches the top number.

$\sum_{0}^{2}( n-1) = (0-1) + (1-1) + (2-1) = -1 + 0 +1 = 0$

Evaluate the next summation.

$\sum_{1}^{3} n = 1+2+3 = 6$

Subtract this value from the value of the first summation.

$\sum_{0}^{2} (n-1 )- \sum_{1}^{3} n = 0-6 = -6$

The answer is:

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Question

Solve: $-\sum_{0}^{2}-n-1$

Answer

To evaluate this summation, first ignore the outside negative sign and substitute the bottom value into the quantity inside the summation. Reiterate for every integer after zero until we reach the top integer.

$\sum_{0}^{2}-n-1 = (-0-1) + (-1-1)+ (-2-1)$

Simplify.

Therefore:

$-\sum_{0}^{2}-n-1 = -(-6) = +6$

The answer is .

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Question

Evaluate: $\sum_{1}^{3}(2n-2)$

Answer

The summation will loop from to .

Rewrite the summation.

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

The answer is:

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Question

Evaluate: $\sum_{0}^{2}(-3n-2)$

Answer

In order to evaluate this summation, expand the terms. Start by plugging in zero into , and repeat until we reach to two.

$\sum_{0}^{2}(-3n-2)=(-3(0)-2)+(-3(1)-2)+(-3(2)-2)$

Simplify the terms by order of operations.

$\sum_{0}^{2}(-3n-2)=-2 +(-3-2)+(-6-2)$

Evaluate the terms inside the parentheses first.

$\sum_{0}^{2}(-3n-2)=-2 -5-8 = -15$

The answer is:

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Question

Evaluate: $\sum_{2}^{4}(6-2n)$

Answer

In order to solve the summation, expand the terms of the binomial. Substitute two first, and add the quantities of each term for each integer repeating until the top integer is reached.

$\sum_{2}^{4}(6-2n)=(6-2(2))+(6-2(3))+(6-2(4))$

The answer is:

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Question

Evaluate: $2\sum_{i=2}^{5}2$

Answer

Write out the terms of the sigma notation. This will repeat for the second to the fifth term. There is no variable to substitute, which means that after every iteration, the value that is summed is two.

$2\sum_{i=2}^{5}2 = 2[2+2+2+2] = 2[8] = 16$

The answer is:

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Question

Evaluate: $\sum_{n=8}^{10} (10-n)$

Answer

In order to evaluate the summation, substitute the bottom number into the quantity, and repeat for every integer after the eight and up to the top number, ten.

$\sum_{8}^{10} (10-n) = (10-8)+(10-9)+(10-10)$

Evaluate each term and simplify.

The answer is:

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Question

Evaluate the summation: $\sum_{n=3}^{6} (1-6n)$

Answer

This summation will repeat three times, for .

Substitute the first term into the parentheses, repeat, and sum the process for the next three terms.

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

Simplify each term.

The answer is:

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Question

Determine the value of: $\sum_{0}^{2}2n^6$

Answer

Expand the summation sign. Start with zero as the first index, then one, and finally two. Since two is the top digit, the summation will stop.

$\sum_{0}^{2}2n^6=2(0)^6+2(1)^6+2(2)^6$

Simplify these terms.

The answer is:

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