For n=1, 2, 3 Tn<2,4,8 respectively is true. So the base case holds.
Assume the hypothesis that Tn≤2^n.
T[n+1]=T[n]+T[n-1]+T[n-2]=T[n-1]+T[n-2]+T[n-3]+T[n-2]+T[n-3]+T[n-4]+T[n-3]+T[n-4]+T[n-5]=
T[n+1]=T[n-1]+2T[n-2]+3T[n-3]+2T[n-4]+T[n-5]
Apply the hypothesis to all T terms on the right.
T[n+1]≤2^(n-1)+2*2^(n-2)+3*2^(n-3)+2*2^(n-4)+2^(n-5)
T[n+1]≤2^(n-1)+2^(n-1)+3*2^(n-3)+2^(n-3)+2^(n-5)
T[n+1]≤2^n+2^(n-1)+2^(n-5)
T[n+1]≤2^(n-5)(32+16+1)
T[n+1]≤2^(n-5)*49
49=2^5.61 approx so 2^(n-5)*49=2^(n-0.61) approx.
T[n+1]≤2^(n-0.61); n+1>n-0.61⇒ 1>-0.61 which is true therefore T[n+1]≤2^(n+1).