<s>[INST] What will be the sum of the series of binomial co-efficients? What will be the sum of the following binomial co-efficent series 
$$\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} = \sum\limits_{i=1}^r \binom{z+i}{z}$$
Thank you
 [/INST] $\dbinom{z+1}{z} +\dbinom{z+2}{z} +\cdots+\dbinom{z+r}{z}$
= Coeff of $x^{z}$ in $(1+x)^{z+1}$+ Coeff of $x^{z}$ in $(1+x)^{z+2}$+$\cdots$+ Coeff of $x^{z}$ in $(1+x)^{z+r}$
= Coeff of $x^{z}$ in $((1+x)^{z+1}+ (1+x)^{z+2}+\cdots+ (1+x)^{z+r})$
= Coeff of $x^{z}$ in $((1+x)^{z+1}\frac{(1+x)^{r}-1}{1+x-1})$
= Coeff of $x^{z+1}$ in $((1+x)^{z+1}.((1+x)^{r}-1))$
= Coeff of $x^{z+1}$ in $((1+x)^{z+r+1}-(1+x)^{z+1})$
=$\dbinom{z+r+1}{z+1}-\dbinom{z+1}{z+1}$
=$\dbinom{z+r+1}{r}-1$
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