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3 years ago

6

Write an expression to describe the sequence below. Use n to represent the position of a term in the sequence, where n = 1 for t

he first term.
6, 7, 8, 9, ...

1 answer:

inessss [21]3 years ago

5 0

Step-by-step explanation:

$a_1=6\\a_2=a_1+1\to a_2=6+1=7\\a_3=a_2+1\to a_3=7+1=8\\a_4=a_3+1\to a_4=8+1=9\\\vdots\\\\\text{The recursive formula}\\a_n=\left\{\begin{array}{ccc}a_1=6\\a_n=a_{n-1}+1\end{array}\right\\\\\text{It's an arithmetic sequence with}\ a_1=6\ \text{and difference}\ d=1\\\\\text{The explicit formula of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=6+(n-1)(1)\\a_n=6+n-1\\a_n=5+n\\\\a_n=n+5$

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3 years ago

Which equation and solution represents the statement shown?

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Answer:

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2 years ago

-(x+3) = 2x-6 what is x

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Answer:

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Step-by-step explanation:

So, first you make sure to multiply the parenthesis by the negative so then it would be,

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2 years ago

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the

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Group Before After

Mean 693.75 743.75

Sd 155.37 143.92

SEM 54.93 50.88

n 8 8

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2 years ago

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

3 years ago