QUESTION 1 ANSWER
The series is
[tex]\sum_{n=1}^{15}{ (2n-1)}[/tex]
So the nth term of this sequence is
[tex]U_n=2n-1[/tex]
We generate some few terms of the sequence as follows:
When [tex]n=1[/tex],
The first term of the sequence becomes,
[tex]U_1=2(1)-1[/tex]
[tex]U_1=1[/tex]
When [tex]n=2[/tex],
[tex]U_2=2(2)-1[/tex]
[tex]U_2=3[/tex]
When [tex]n=3[/tex],
[tex]U_2=2(3)-1[/tex]
[tex]U_2=5[/tex]
The constant difference is [tex]d=2[/tex].
The sum of the series is given by,
[tex]S_n=\frac{n}{2}(2a+(n-1)d)[/tex]
There are 15 terms in the series, therefore [tex]n=15[/tex]
[tex]S_{15}=\frac{15}{2}(2(1)+(15-1)2)[/tex]
[tex]S_{15}=\frac{15}{2}(2+14(2))[/tex]
[tex]S_{15}=\frac{15}{2}(30)[/tex]
[tex]S_{15}=15\times 15[/tex]
[tex]S_{15}=225[/tex]
ANSWER TO QUESTION 2
The bottom row has [tex]23[/tex] boxes. This means [tex]a=U_1=23[/tex]
a) Since each row has 3 fewer boxes [tex]d=-3[/tex].
The nth row is given by,
[tex]U_n=23+(n-1)\times(-3)[/tex]
[tex]U_n=23+-3n+3[/tex]
[tex]U_n=23-3n[/tex]
The top row is the 6th row, meaning [tex]n=6[/tex].
[tex]U_6=23-3(6)[/tex]
[tex]U_6=23-18[/tex]
[tex]U_6=5[/tex]
b) The number of boxes in the entire display is given by
[tex]S_n=\frac{n}{2} (a+l)[/tex]
Where [tex]a=23[/tex], the first term(row) and [tex]l=5[/tex], the last term(row).
Since there are 6 rows,
[tex]S_6=\frac{6}{2} (23+5)[/tex]
[tex]S_6=3 (28)[/tex]
[tex]S_6=84[/tex]
ANSWER TO QUESTION 3
The number of visitors in the first week is [tex]a=5[/tex].
Since the number of visitors each week is doubled the number of visitors in the subsequent weeks, the sequence is a geometric progression with a common ratio of [tex]r=2[/tex].
The general term of a geometric sequence is
[tex]U_n=ar^{n-1}[/tex]
So in the 8th week, we have
[tex]U_8=5\times 2^{8-1}[/tex]
[tex]U_8=5\times 2^{7}[/tex]
[tex]U_8=5\times 128[/tex]
[tex]U_8=640[/tex]
Hence 640 people visited the website in the 8th week
