Sequences and Series - High School Math

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Question

Find the sum of all even integers from to .

Answer

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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Question

Find the sum of all even integers from to .

Answer

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 and the last term is . Our formula becomes:

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

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Question

Find the sum of the even integers from to .

Answer

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 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:

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

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Question

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

Answer

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

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Question

Evaluate:

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

Answer

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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Question

Evaluate:

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

Answer

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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Question

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

What is the fifteenth term in the sequence?

Answer

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

For the 15th term, .

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

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Question

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

Answer

We will need to use the Binomial Theorem in order to solve this problem. Consider the expansion of , 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 . 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 $\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$ .

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

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

The answer is .

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Question

What are the first three terms in the series?

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

Answer

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

The first three terms are , , and .

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Question

Find the first three terms in the series.

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

Answer

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

The first three terms are , , and .

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Question

Indicate the first three terms of the following series:

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

Answer

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

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Question

Indicate the first three terms of the following series:

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

Answer

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

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Question

Indicate the first three terms of the following series:

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

Answer

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

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

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

$(2)^{0}=1$

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Question

Indicate the first three terms of the following series.

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

Answer

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

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

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

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

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Question

Evaluate: $1 - \frac{2}{3} + \frac{4}{9} - \frac{8}{27} + ...$

Answer

This sum can be determined using the formula for the sum of an infinite geometric series, with initial term $a_{0} = 1$ and common ratio $r = -\frac{2}{3}$ :

$S = \frac{a_{0}} {1-r} = \frac{1} {1- \left ( -\frac{2}{3}\right ) } = \frac{1} {1+ \frac{2}{3} }= \frac{1} { \frac{5}{3} } =\frac{3}{5}$

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Question

The fourth term in an arithmetic sequence is -20, and the eighth term is -10. What is the hundredth term in the sequence?

Answer

An arithmetic sequence is one in which there is a common difference between consecutive terms. For example, the sequence {2, 5, 8, 11} is an arithmetic sequence, because each term can be found by adding three to the term before it.

Let $a_{n}$ denote the nth term of the sequence. Then the following formula can be used for arithmetic sequences in general:

, where d is the common difference between two consecutive terms.

We are given the 4th and 8th terms in the sequence, so we can write the following equations:

.

We now have a system of two equations with two unknowns:

Let us solve this system by subtracting the equation from the equation . The result of this subtraction is

.

This means that d = 2.5.

Using the equation , we can find the first term of the sequence.

Ultimately, we are asked to find the hundredth term of the sequence.

$a_{100}=a_1+(100-1)d=-27.5+99(2.5)=220$

The answer is 220.

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Question

Find the sum, if possible:

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

Answer

The formula for the summation of an infinite geometric series is

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

where is the first term in the series and is the rate of change between succesive terms. The key here is finding the rate, or pattern, between the terms. Because this is a geometric sequence, the rate is the constant by which each new term is multiplied.

Plugging in our values, we get:

$S = \frac{3}{1-\frac{1}{3}}$

$S=\frac{3}{\frac{2}{3}}$

$S = \frac{9}{2}$

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Question

Find the sum, if possible:

$8-6+\frac{9}{2}-\frac{27}{8}+...$

Answer

The formula for the summation of an infinite geometric series is

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

where is the first term in the series and is the rate of change between succesive terms in a series

Because the terms switch sign, we know that the rate must be negative.

Plugging in our values, we get:

$S = \frac{8}{1-(\frac{-3}{4})}$

$S = \frac{8}{\frac{7}{4}}$

$S = \frac{32}{7}$

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Question

Find the sum, if possible:

Answer

The formula for the summation of an infinite geometric series is

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

where is the first term in the series and is the rate of change between succesive terms in a series.

In order for an infinite geometric series to have a sum, needs to be greater than and less than , i.e. .

Since , there is no solution.

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Question

Determine the summation notation for the following series:

Answer

The series is a geometric series. The summation notation of a geometric series is

$\sum_{n=1}^{n}a_{1}\cdot r^{n-1}$ ,

where is the number of terms in the series, $a_{1}$ is the first term of the series, and is the common ratio between terms.

In this series, is , $a_{1}$ is , and is . Therefore, the summation notation of this geometric series is:

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

This simplifies to:

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

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