Insertion Method:
Total number of nodes present present in a binary tree at level $i$ is $2^{i-1}$
Number of comparisons involved for inserting a node at level $i=i-1$
Number of comparisons of all nodes at level $i=(i-1)2^{i-1}$
$\begin{array}{ll} \text{Total number of comparisons } &=\Sigma_{i=1}^k (i-1) 2^{i-1} \\ & = (k-2)2^k +2 \\ & \bigg[ \Sigma_{i=1}^n (i)2^i = (k-1)2^{k+1}+2 \bigg] \\ & = k2^k -2^{k+1}+2 \end{array}$
