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enot [183]
3 years ago
7

Show work explain with formulas.

Mathematics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}

1. a

44

=481,

n=1

∑

∞

(11n−3), the sum not exist

\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}

3.

n=1

∑

47

(9n+16)=10,904

\large\boxed{4.\ \sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg),\ \text{the sum not exist}}

4.

n=1

∑

∞

(6(2

n

)), the sum not exist

Step-by-step explanation:

\begin{lgathered}1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481\end{lgathered}

1.

8+19+30+41+...

19−8=11

30−19=11

41−30=11

It’s an arithmetic series. a

1

=8, d=11.

The formula for the n-th term of an arithmetic sequence:

a

n

=a

1

+(n−1)d

Substitute:

a

n

=8+(n−1)(11)use the distributive property a(b−c)=ab−ac

a

n

=8+11n−11

a

n

=11n−3

Calculate the 44th term. Put n = 44 to the formula:

a

44

=(11)(44)−3=484−3=481

\begin{lgathered}\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.\end{lgathered}

The summation notation:

n=1

∑

∞

(11n−3)

The sum not exist, because d>1, therefore 11n−3→∞.

=======================================================

\begin{lgathered}3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47\end{lgathered}

3.

25+34+43+52+...+436

34−25=9

43−34=9

52−43=9

It’s an arithmetic series. a

1

=25, d=9.

The formula for the n-th term of an arithmetic sequence:

a

n

=a

1

+(n−1)d

Substitute:

a

n

=25+(n−1)(9)

a

n

=25+9n−9

a

n

=9n+16

Calculate which the term of arithmetic series is the number 439.

Put a

n

=439:

9n+16=439subtract 16 from both sides

9n=423divide both sides by 9

n=47

\begin{lgathered}\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904\end{lgathered}

Therefore we have the series in summation notation:

n=1

∑

47

(9n+16)

For calculation of the sum we use the formula of a sum of terms

of an arithmetic sequence:

S

n

=

2

a

1

+a

n

⋅n

Substitute n=47, a

1

=25, a

47

=439:

S

47

=

2

25+439

⋅47=

2

464

⋅47=232⋅47=10904

n=1

∑

47

(9n+16)=10,904

=======================================================

\begin{lgathered}4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty\end{lgathered}

4.

12+24+48+...

24:12=2

48:24=2

It’s a geometric series a

1

=12, r=2.

The formula for the n-th term of a geometic sequence:

a

n

=a

1

r

n−1

Substitute:

a

n

=(12)(2

n−1

)use

a

m

a

n

=a

n−m

a

n

=(12)(

2

1

2

n

)

a

n

=(6)(2

n

)

The summation notation:

n=1

∑

∞

(6(2

n

))

The sum not exist, because r>1, therefore 6(2

n

)→∞

Sedaia [141]3 years ago
6 0

Answer:

\large\boxed{1.\ a_{44}=481,\ \sum\limits_{n=1}^\infty(11n-3),\ \text{the sum not exist}}

\large\boxed{3.\ \sum\limits_{n=1}^{47}(9n+16)=10,904}

\large\boxed{4.\ \sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg),\ \text{the sum not exist}}

Step-by-step explanation:

1.\\8+19+30+41+...\\\\19-8=11\\30-19=11\\41-30=11\\\\\text{It's an arithmetic series.}\ a_1=8,\ d=11.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=8+(n-1)(11)\qquad\text{use the distributive property}\ a(b-c)=ab-ac\\a_n=8+11n-11\\\boxed{a_n=11n-3}\\\\\text{Calculate the 44th term. Put n = 44 to the formula:}\\\\a_{44}=(11)(44)-3=484-3=481

\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty(11n-3)\\\\\text{The sum not exist, because}\ d>1,\ \text{therefore}\ 11n-3\rightarrow\infty.

=======================================================

3.\\25+34+43+52+...+436\\\\34-25=9\\43-34=9\\52-43=9\\\\\text{It's an arithmetic series.}\ a_1=25,\ d=9.\\\\\text{The formula for the n-th term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\\text{Substitute:}\\\\a_n=25+(n-1)(9)\\a_n=25+9n-9\\a_n=9n+16\\\\\text{Calculate which the term of arithmetic series is the number 439.}\\\text{Put}\ a_n=439:\\\\9n+16=439\qquad\text{subtract 16 from both sides}\\9n=423\qquad\text{divide both sides by 9}\\n=47

\text{Therefore we have the series in summation notation:}\\\\\sum\limits_{n=1}^{47}(9n+16)\\\\\text{For calculation of the sum we use the formula of a sum of terms}\\\text{of an arithmetic sequence:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\text{Substitute}\ n=47,\ a_1=25,\ a_{47}=439:\\\\S_{47}=\dfrac{25+439}{2}\cdot47=\dfrac{464}{2}\cdot47=232\cdot47=10904\\\\\sum\limits_{n=1}^{47}(9n+16)=10,904

=======================================================

4.\\12+24+48+...\\\\24:12=2\\48:24=2\\\\\text{It's\ a\ geometric series}\ a_1=12,\ r=2.\\\\\text{The formula for the n-th term of a geometic sequence:}\\\\a_n=a_1r^{n-1}\\\\\text{Substitute:}\\\\a_n=(12)(2^{n-1})\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\a_n=(12)\left(\dfrac{2^n}{2^1}\right)\\\\a_n=(6)(2^n)\\\\\text{The summation notation:}\\\\\sum\limits_{n=1}^\infty\bigg(6(2^n)\bigg)\\\\\text{The sum not exist, because}\ r>1,\ \text{therefore}\ 6(2^n)\to\infty

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