Answer:
4294967293 dots
Step-by-step explanation:
Figure 1 has 5 dots.
Figure 2 has 13 dots.
Figure 3 has 29 dots.
Subtract and find different of each term:
Let [tex]\displaystyle \large{b_n}[/tex] be sequence of difference (and the sequence appears to be geometric.)
So for [tex]\displaystyle \large{b_n}[/tex], find the common ratio which is 2.
Hence, [tex]\displaystyle \large{b_n = 8(2)^{n-1}}[/tex] - recall geometric sequence formula below:
[tex]\displaystyle \large{a_n = a_1r^{n-1}}[/tex]
The sequence of difference [tex]\displaystyle \large{b_n=8(2)^{n-1}}[/tex] can be simplified to:
[tex]\displaystyle \large{b_n=2^3(2)^{n-1}}\\\displaystyle \large{b_n=2^{n+2}}[/tex]
Now to find the original sequence:
[tex]\displaystyle \large{a_n = a_1 + \sum_{k=1}^{n-1}b_k}[/tex]
Hence:
[tex]\displaystyle \large{T_n=5+\sum_{k=1}^{n-1}8(2)^{k-1}}[/tex]
Recall:
[tex]\displaystyle \large{\sum_{k=1}^{n-1} a_1r^{n-1} =a_1\left( \dfrac{1-r^{n-1}}{1-r}\right)}[/tex]
Therefore:
[tex]\displaystyle \large{T_n=5+\sum_{k=1}^{n-1}8(2)^{k-1}}\\\displaystyle \large{T_n=5+8\left(\dfrac{1-2^{n-1}}{1-2}\right)}\\\displaystyle \large{T_n=5+8\left(\dfrac{1-2^{n-1}}{-1}\right)}\\\\\displaystyle \large{T_n=5+8(-1+2^{n-1})}\\\displaystyle \large{T_n=5-8+8(2)^{n-1}}\\\displaystyle \large{T_n=2^{n+2}-3}[/tex]
Therefore, the sequence for 5,13,29 is [tex]\displaystyle \large{2^{n+2}-3}[/tex].
Therefore, in figure 30:
[tex]\displaystyle \large{T_{30}=2^{30+2}-3}\\\displaystyle \large{T_{30}=2^{32}-3}\\\displaystyle \large{T_{30}=4294967293}[/tex]
Therefore, there are 4294967293 dots in figure 30
