Solution :
Let the base case, n = 0
Given at
at least
computers are infected
So,

Therefore, n = 0 is not true
Let suppose, it is true for n = k
Then,

We have to show that it is also true for

That is at the end of (k+1) second,

Now, the number of the infected computer that is added between
and
second =
![$(1)/(5)\left[50+12(0.6)^k\right]$](https://img.qammunity.org/2022/formulas/mathematics/college/aiy0mgat6phfrppenwadwejhws4spchd36.png)
Now the number of the infected computer decreased between
and
second is = 10.
Hence at the end of the
second.
Number of the computers
![$\geq 50+12(0.6)^k + 1/5[50+12(0.6)^k]-10$](https://img.qammunity.org/2022/formulas/mathematics/college/vhgkvyg2nzaor04d1my5mer4uyr6bn4yej.png)
![$\geq 50+12(0.6)^k +10+ 0.5[12(0.6)^k]-10$](https://img.qammunity.org/2022/formulas/mathematics/college/kxytm7lq134hpfp6ddf1yxhlfmn3ra9ivh.png)
![$\geq 50+12(0.6)^k+0.5[12(0.6)^k]$](https://img.qammunity.org/2022/formulas/mathematics/college/kghwm5tat2fwi415hkui9oordpee4fksi5.png)


Hence Proved