If n= 4, then there will be four terms in the sum, for i= 1, 2, 3, and 4. For i= 1 that term will be $\displaystyle K_1(1+ 2H_{14}K_1+ 2H_{24}K_2+ 2H_{34}K_3+ 2H_{44}K_4+ H_{22}K_2^2+ H_{23}K_2K_3+ H_{24}K_2K_4+ H_{32}K_3K_2+ H_{42}K_4K_3+ H_{13}K_1K_3+ H_{23}K_2K_3+ H_{33}K_3^3+ H_{43}K_4K_3+ H_{14}K_1K_4+ H_{24}K_2K_4+ H_{34}K_3K_4+ H_{44}K_4^2$
The i= 2 term will be similar and so will the i= 3 and i= 4 terms.