Please help to show the following identity

Question

How can I prove the following identity ? $$ k \binom{n}{k} \sum_{i=0}^{k-1} \frac{\binom{k-1}{i}(-1)^i}{(n-k+1+i)} e^{-\frac{T}{\theta_1}(n-k+1+i)}=\sum_{i=0}^{k-1} \binom{n}{i} (1-e^{-\frac{T}{\theta_1}})^ie^{-\frac{T}{\theta_1}(n-i)}. $$

Answer 1

Let $z=e^{\frac{T}{\theta_1}}$. Then the RHS is $$z^{-n}\sum_{i=0}^{k-1} \binom{n}{i} (z-1)^i=z^{-n}\sum_{i=0}^{k-1} \binom{n}{i} \sum_{j=0}^i(-1)^{i-j}\binom{i}{j} z^{j}= z^{-n}\sum_{j=0}^{k-1}(-1)^jz^{j}\sum_{i=j}^{k-1}(-1)^{i}\binom{n}{i}\binom{i}{j}.$$ Now the inner sum gives $$\frac{(-1)^{k-1} k}{n-j}\binom{n}{k}\binom{k-1}{j}.$$ Therefore $$\mbox{RHS}=z^{-n}k\binom{n}{k}\sum_{j=0}^{k-1}\frac{(-1)^{j-(k-1)}}{n-j}\binom{k-1}{j}z^{j}=z^{-n}k\binom{n}{k}\sum_{j=0}^{k-1}\frac{(-1)^{j-(k-1)}}{n-j}\binom{k-1}{k-1-j}z^{j}\\ =k\binom{n}{k}\sum_{i=0}^{k-1}\frac{(-1)^{i}}{n-(k-1-i)}\binom{k-1}{i}z^{k-1-i-n} $$ which is equal to the LHS.