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$(1+x)^b=\sum_{k=0}^{b} x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\color{cadetblue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$

$(1+x)^b=\sum_{k=0}^{b} \binom{b}{k}x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\color{cadetblue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$

$(1+x)^b=\sum_{k=0}^{b} x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\color{cadetblue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$

$(1+x)^b=\sum_{k=0}^{b} x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\color{cadetblue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$

$(1+x)^b=\sum_{k=0}^{b} x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\texcolor{Blue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$

$(1+x)^b=\sum_{k=0}^{b} x^k$

Now, $\binom{a+b-1-i}{a-i}=\binom{a+b-1-i}{b-1}$...$(1)$

And, $(1+x)^{-b}=1-\binom{b}{b-1}x+\binom{b+1}{b-1}x^2-..\color{cadetblue}{(-1)^{a-b}\binom{a-1}{b-1}x^{a-b}+(-1)^{a-b+1}\binom{a}{b-1}x^{a-b+1}......+(-1)^{a}\binom{a+b-1}{b-1}x^{a+1}}+.... \tag{2}$

Multiplying (1) and (2) we easily see that the coefficient of $x^a$ in r.h.s $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$.

But the left hand side of $(1)×(2)$ is 1. Hence, the coefficient of $x^a, a\geq 1$ is $0$.

Hence, $$(-1)^a\sum_{I=0}^{b}(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}=0$$