Help in Linear summation of n(any given number)

Question

The original formula is this.

We're computing the complexity of an insertion sort.

How did the first formula turned into the second formula?

Answer 1

$t_j = 1$ for $j = 2,\ldots,n$, so the sums with $(t_j-1)$ vanish, and $\sum_{j=2}^{n} t_j = n-1$. We then get:

$$\begin{align} T(n) &= c_1 n + c_2(n-1) + c_4(n-1) + c_5(n-1) + c_8(n-1) \\ &= c_1 n + (c_2 + c_4 + c_5 + c_8)(n-1) \\&= c_1 n + (c_2 + c_4 + c_5 + c_8)n - (c_2 + c_4 + c_5 + c_8) \\&= (c_1 + c_2 + c_4 + c_5 + c_8)n - (c_2 + c_4 + c_5 + c_8) \end{align} $$