We can see least number of nodes will be created from T(n/2).
Thus T(n/2) defines lower bound of T(n)
Sum of outputs by T(n/2) = n/2+n/4+n/8+.........= n(1/2+1/4+1/8+...)
= n[1/(1-1/2)] =2n
2n<=T(n)
Similarly
maximum number of nodes will be created from T(n/8).
Thus T(n/8) defines upper bound of T(n)
Sum of outputs by T(n/8) = 7n/8+7n/8 2 +7 n/8 3+.........= = n[1/(1-7/8)] =8n
T(n)<=8n
We conclude 2n <= T(n) <= 8n
T(n) = θ(n)