1. Define a sequence of numbers in the following way: a[k]=a[k-1]+a[k-2]+a[k-3] for k≥3, s.t. a=1, a=2, a=3, a=6... *The numbers in brackets will be subscripts for the whole problem Prove that a[n]≤2^n using complete mathematical induction. This is what I have so far. We'll use complete mathematical induction, where P(n)=a[n]≤2^n P(0) is the statement a≤2^0. This is true because 1≤1. Assume For all n, 0, 1, ..., k a[n]≤2^n is true. We want to show that for k+t that a[k+1]≤2^(k+1) Thus, a[k+1]=a[k]+a[k-1]+a[k-2] And 2^(k+1)=2(2^k) Therefore, a[k]+a[k-1]+a[k-2]≤2(2^k). I can't seem to get past this part. Any help would be greatly appreciated.