Question

1. Find the sum of the series (IMAGE) . Show your work.

Answer 1

QUESTION 1 ANSWER

The series is

$\sum_{n=1}^{15}{ (2n-1)}$

So the nth term of this sequence is

$U_n=2n-1$

We generate some few terms of the sequence as follows:

When $n=1$,

The first term of the sequence becomes,

$U_1=2(1)-1$

$U_1=1$

When $n=2$,

$U_2=2(2)-1$

$U_2=3$

When $n=3$,

$U_2=2(3)-1$

$U_2=5$

The constant difference is $d=2$.

The sum of the series is given by,

$S_n=\frac{n}{2}(2a+(n-1)d)$

There are 15 terms in the series, therefore $n=15$

 $S_{15}=\frac{15}{2}(2(1)+(15-1)2)$

$S_{15}=\frac{15}{2}(2+14(2))$

$S_{15}=\frac{15}{2}(30)$

$S_{15}=15\times 15$

$S_{15}=225$

ANSWER TO QUESTION 2

The bottom row has $23$ boxes. This means $a=U_1=23$

a) Since each row has 3 fewer boxes $d=-3$.

The nth row is given by,

$U_n=23+(n-1)\times(-3)$

$U_n=23+-3n+3$

$U_n=23-3n$

The top row is the 6th row, meaning $n=6$.

$U_6=23-3(6)$

$U_6=23-18$

$U_6=5$

b) The number of boxes in the entire display is given by

$S_n=\frac{n}{2} (a+l)$

Where $a=23$, the first term(row) and $l=5$, the last term(row).

Since there are 6 rows,

$S_6=\frac{6}{2} (23+5)$

$S_6=3 (28)$

$S_6=84$

ANSWER TO QUESTION 3

The number of visitors in the first week is $a=5$.

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 $r=2$.

The general term of a geometric sequence is

$U_n=ar^{n-1}$

So in the 8th week, we have

$U_8=5\times 2^{8-1}$

$U_8=5\times 2^{7}$

$U_8=5\times 128$

$U_8=640$

Hence 640 people visited the website in the 8th week