Finding Terms in a Series - Math

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Indicate the first three terms of the following series:

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

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

Indicate the first three terms of the following series:

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

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

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

What is the fifteenth term in the sequence?

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

For the 15th term, .

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

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

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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:

,

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 .

What are the first three terms in the series?

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

Tap to see back →

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

The first three terms are , , and .

Find the first three terms in the series.

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

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

The first three terms are , , and .

Indicate the first three terms of the following series:

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

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

Indicate the first three terms of the following series.

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

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

Indicate the first three terms of the following series:

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

Tap to see back →

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

Indicate the first three terms of the following series:

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

Tap to see back →

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

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

What is the fifteenth term in the sequence?

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

For the 15th term, .

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

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

Tap to see back →

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:

,

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 .

What are the first three terms in the series?

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

Tap to see back →

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

The first three terms are , , and .

Find the first three terms in the series.

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

Tap to see back →

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

The first three terms are , , and .

Indicate the first three terms of the following series:

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

Tap to see back →

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

Indicate the first three terms of the following series.

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

Tap to see back →

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

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

What is the fifteenth term in the sequence?

Tap to see back →

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

For the 15th term, .

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

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

Tap to see back →

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:

,

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 .

What are the first three terms in the series?

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

Tap to see back →

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

The first three terms are , , and .

Find the first three terms in the series.

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

Tap to see back →

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

The first three terms are , , and .