How do you do part (c)? Is there a shorter way than calculating everything upto 100?
$\displaystyle A_2=A_1+2(1)$
$\displaystyle A_3=A_2+2(2)$
$\displaystyle A_4=A_3+2(3)$
$\displaystyle \vdots$
$\displaystyle A_{n-1}=A_{n-2}+2(n-2)$
$\displaystyle A_{n}=A_{n-1}+2(n-1)$
add downwards, which eventually gives
$\displaystyle A_n=A_1+2(1+2+3+...+n-1)$
take$\displaystyle A_1=4$
$\displaystyle A_n=4+2\sum_{r=1}^{n-1} r$
Hello, fleurdel!
How do you do part (c)?
Is there a shorter way than calculating everything up to 100?
(c) Find the area of Figure 100.
. . $\displaystyle \begin{array}{cc} \text{Figure} & \text{Area} \\ \hline 1 & 6 \\ 2 & 10 \\ 3 & 16 \\ 4 & 24 \\ \vdots & \vdots \end{array}$
We have this sequence: .$\displaystyle 6,\,10,\,16,\,24\, \hdots$
Take the difference of consecutive terms,
. . then take the difference of the differences, and so on.
. . $\displaystyle \begin{array}{cccccccc}\text{Sequence} & 6 && 10 && 16 && 24 \\ \text{1st diff.} && 4 && 6 && 8 \\ \text{2nd diff.} &&& 2 && 2 \end{array}$
We see that the second differences are constant.
This mean that the generating function is of the second degree . . . a quadratic.
The general quadratic function is: $\displaystyle f(n) \,=\,an^2 + bn + c$
Use the first three terms of the sequence to construct a system of equations:
. . $\displaystyle \begin{array}{cccccc}f(1) = 6 & a + b + c &=& 6 & [1] \\ f(2) = 6 & 4a + 2b + c &=& 10 & [2] \\ f(3) = 16 & 9a + 3b + c &=& 16 & [3] \end{array}$
. . $\displaystyle \begin{array}{cccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 4 & [4] \\ \text{Subtract [3] - [2]:} & 5a + b &=& 6 & [5] \end{array}$
. . $\displaystyle \text{Subtract [5] - [4]: }\;2a \:=\:2 \quad\Rightarrow\quad a \:=\:1$
$\displaystyle \text{Substitute into [4]: }\:3 + b \:=\:4 \quad\Rightarrow\quad b \:=\:1$
$\displaystyle \text{Substitute into [1]: }\; 1 + 1 + c \:=\:6 \quad\Rightarrow\quad c \:=\:4$
. . $\displaystyle \text{Hence: }\:f(n) \:=\:n^2 + n + 4$
$\displaystyle \text{Therefore: }\:f(100) \:=\:100^2 + 100 + 4 \:=\:\boxed{10,\!104}$