How to find the value of the following items summed up together?

Question

How to find the value of the following items summed up together?

$$U^j_nd^0_n\binom{j}{j}P^j_n(1-P_n)^0+U^{j-1}_nd^1_n\binom{j}{j-1}P^{j-1}_n(1-P_n)^1+U^{j-2}_nd^2_n\binom{j}{j-2}P^{j-2}_n(1-P_n)^2+\dots+U^2_nd^{j-2}_n\binom{j}{2}P^{2}_n(1-P_n)^{j-2}+U^1_nd^{j-1}_n\binom{j}{1}P^{1}_n(1-P_n)^{j-1}+U^0_nd^{j}_n\binom{j}{0}P^{0}_n(1-P_n)^{j}$$

Answer 1

This appears to be a direct application of the binomial theorem for $(P_nU_n + (1-P_n)d_n)^j$

Binomial theorem: $$(x+y)^n = \sum\limits_{i=0}^n \binom{n}{i}x^iy^{n-i}$$

To see that, group the terms with like-powers. (For the term $U_n^{j-1}d_n^1\binom{j}{j-1}P_n^{j-1}(1-P_n)^1$ for example, put the $U$ and the $P$ term together and the $d$ and the $(1-P)$ term together). You get a sum of powers with one set increasing in power, the other decreasing in power with a binomial coefficient on each, matching the statement of the theorem.

Setting $x= U_nP_n$, $y=d_n(1-P_n)$, and $n=j$ into the above formula gives the result.