What will be the summation of this Series
$$\binom{10}{1} + \binom{11}{2} + \binom{12}{3} +\cdots+\binom{10+n}{n+1}$$
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See $\sum_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}$ – Bart Michels Apr 08 '15 at 15:07
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Note that this is the same as $$\binom {10}9+\binom{11}9+\binom{12}9+\cdots+\binom{10+n}9$$ or $$\sum_{r=0}^{n}\binom{r+10}9$$ or $$\sum_{r=1}^{n+1}\binom{r+9}9$$ – Hypergeometricx Apr 08 '15 at 15:33
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If you add $\dbinom{10}0$ to your sum and apply successively the Pascal's identity you have: $$\dbinom{n}k+\dbinom{n}{k+1}=\dbinom{n+1}{k+1}$$ you will get:
$$\underbrace{\binom{10}{0} + \binom{10}{1}}_{\dbinom{11}{1}} + \binom{11}{2} +\cdots+\binom{10+n}{n+1}$$
and again $\dbinom{11}{1}$ will concel with $\dbinom{11}{1}$ to get $\dbinom{12}{2}$ and so on$\cdots$. The sum equals in the end: $$\binom{11+n}{n+1}-\binom{10}{0}=\binom{11+n}{10}-1$$
Elaqqad
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People keep asking the same question over and over these days, maybe it is time to add an answer as a reference for forthcoming identical questions.
It is straightforward to prove by induction that:
$$\sum_{k=0}^{K}\binom{n+k}{n}=\binom{n+K+1}{n+1}\tag{1}$$ hence your sum equals: $$ \sum_{k=1}^{n+1}\binom{9+k}{9}=\binom{11+n}{10}-1.\tag{2}$$
Jack D'Aurizio
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