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
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}\\\=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
)→∞