Respuesta :
Solution :
Let the base case, n = 0
Given at [tex]$t=0,$[/tex] at least [tex]60[/tex] computers are infected
So, [tex]$12\times \left(\frac{6}{10}\right)^0+50=60$[/tex]
Therefore, n = 0 is not true
Let suppose, it is true for n = k
Then, [tex]$C(k) \geq 50+12(0.6)^k$[/tex]
We have to show that it is also true for [tex]$n=(k+1)$[/tex]
That is at the end of (k+1) second,
[tex]$C(k+1) \geq 50+12 \times (0.6)^{k+1}$[/tex]
Now, the number of the infected computer that is added between [tex]n=k[/tex] and [tex]$n=(k+1)$[/tex] second = [tex]$\frac{1}{5}\left[50+12(0.6)^k\right]$[/tex]
Now the number of the infected computer decreased between [tex]n=k[/tex] and [tex]$n=(k+1)$[/tex] second is = 10.
Hence at the end of the [tex]$(k+1)$[/tex] second.
Number of the computers [tex]$\geq 50+12(0.6)^k + 1/5[50+12(0.6)^k]-10$[/tex]
[tex]$\geq 50+12(0.6)^k +10+ 0.5[12(0.6)^k]-10$[/tex]
[tex]$\geq 50+12(0.6)^k+0.5[12(0.6)^k]$[/tex]
[tex]$\geq 50+12(0.6)^k+2(0.6)^k$[/tex]
[tex]$ \geq 50+12 \times (0.6)^{k+1}$[/tex]
Hence Proved
