I believe you are referring to Trojan.SpyAxe32.Downloader [Symantec]
Problem 1:
Define the binomial coefficient
n by
r
n = n! / (r!(n-r)!) for r = 0,1,2,...., n
r
problem: show that
n
r
+
n
r-1
=
n+1
r
for r = 0,1,2,...,n
problem 2:
Use induction to prove Bernoulli's inequality:
If 1+x>0, then (1+x)^n >= 1+nx for all n (element) of N
N is the set of positive integers, natural numbers
I believe you are referring to Trojan.SpyAxe32.Downloader [Symantec]
For problem 2:
If 1 + x > 0 then (1 + x)^n >= 1 + nx for all n.
Since 1 + x > 0, x > -1. So n*x^2 is nonnegative for all n in N and x in R.
Now we show by induction that (1 + x)^n >= 1 + nx for all n
Let P(n) be (1 + x)^n >= 1 + nx for all n.
=> P(1): (1 + x)^1 >= 1 + 1*x, which is true.
Since P(1) is true, assume P(n) is true. We show P(n) => P(n+1)
Now (1 + x)^(n + 1) = (1 + x)*(1 + x)^n
>= (1 + x)(1 + nx)
= 1 + x + nx + nx^2
= 1 + (1 + n)x + nx^2
>= 1 + (1 + n)x since nx^2 is nonnegative.
Thus we have (1 + x)^(n + 1) >= 1 + (n + 1)x which is P(n + 1). So since P(n + 1) is true, we can conclude that P(n) is true for all n. That is, (1 + x)^n >= 1 + nx for all n
Yeah, it's always the same principle, (with some variations for more complicated problems).
You have a statement P(n) you want to prove for all n. You begin by showing P(1) is true. If it is, you assume P(n) is true, and then show that P(n + 1) is true as a consequence of P(n) being true.
Variations are usually in the first step. For example, for a class i had last year, i had to show up to P(5) was true before i could solve the problem.
Thanks for the well wishes, i'll need them. good luck with your classes as well