Answer:
15. The numbers 3,-6, 12, -24,...
16. The numbers -3, -1,5, 23,...
17, The numbers 3, 9, 15, 21,...
18. [tex]\sum_{n=1}^{\infty} 4(\frac{1}{2})^{n-1}[/tex]
19. The sum is 8
20. [tex]a_{10}=-12[/tex]
Step-by-step explanation:
Question 15:
For the recursive sequence [tex]a_1=3, a_n=a_{n-1}*(-2)[/tex]
[tex]a_1=3,\\a_2=3(-2)=-6\\a_3=-6(-2)=12\\a_4=12(-2)=-24[/tex]
Therefore we get the numbers: 3, -6, 12, -24, ...
Question 16:
For the sequence [tex]a_n=3^{n-1}-4[/tex]
[tex]a_1=-3\\a_2=-1\\a_3=5\\a_4=23[/tex]
Therefore we have the numbers -3, -1, 5, 23, ....
Question 17.
For the recursive equation [tex]a_n=a_{n-1}+6[/tex]
[tex]a_1=3[/tex]
[tex]a_2=3+6=9[/tex]
[tex]a_3=9+6=15[/tex]
[tex]a_4=15+6=21[/tex]
The numbers generated are 3, 9, 15, 21,...
Question 18.
In the series [tex](4,2,1,\frac{1}{2},... )[/tex]
each following number is [tex]\frac{1}{2}[/tex] the previous number. The first term is 4, and the sequence has infinite terms; therefore, we have
[tex]\sum_{n=1}^{\infty} 4(\frac{1}{2})^{n-1}[/tex]
Question 19.
[tex]\sum_{n=1}^{\infty} 4(\frac{1}{2})^{n-1}=4*\sum_{n=1}^{\infty} (\frac{1}{2})^{n-1}[/tex]
since
[tex]\sum_{n=1}^{\infty} (\frac{1}{2})^{n-1}=1-\frac{1}{2^{n-1}}[/tex]
as n approaches infinity
[tex]\lim_{n \to \infty} 1-\frac{1}{2^{n-1}}=1[/tex]
therefore we have
[tex]\sum_{n=1}^{\infty} 4(\frac{1}{2})^{n-1}=4*\sum_{n=1}^{\infty} (\frac{1}{2})^{n-1}=4*1=4[/tex]
[tex]\boxed{\sum_{n=1}^{\infty} 4(\frac{1}{2})^{n-1}=4}[/tex]
The sum is 4.
Question 20.
an arithmetic sequence has the form
[tex]a_n=a_1+(n-1)d[/tex]
In our case [tex]a_1=6[/tex] and [tex]d=-2[/tex]; therefore we have
[tex]a_n=6-2(n-1)[/tex]
The 10th term of this sequence would be
[tex]a_{10}=6-2(10-1)=6-2(9)=-12[/tex]
[tex]\boxed{a_{10}=-12}[/tex]
