Sums of Sums of Sums of

Let $f(n,m)=\underbrace{\displaystyle\sum^{m}_{a_1=1}\sum^{a_1}_{a_2=1}\cdots \sum^{a_{n-1}}_{a_{n}=1}}_{n\text{ sums}}1$ We shall prove using induction that $f(n,m)=\dbinom{n+m-1}{n}$ .

Base case: $n=1$

Clearly, $f(1,m)=\displaystyle\sum^{m}_{a_1=1}1=\dbinom{m}{1}$ .

Induction step:

Let us assume $f(k,m)=\underbrace{\displaystyle\sum^{m}_{a_1=1}\sum^{a_1}_{a_2=1}\cdots \sum^{a_{k-1}}_{a_{k}=1}}_{i\text{ sums}}1=\dbinom{k+m-1}{k}$

We can see that $f(k+1,m)=\underbrace{\displaystyle\sum^{m}_{a_1=1}\sum^{a_1}_{a_2=1}\cdots \sum^{a_{k}}_{a_{k+1}=1}}_{k+1\text{ sums}}1=\displaystyle\sum^{m}_{i=1}f(k,i)=\displaystyle\sum^{m}_{i=1}\dbinom{k+i-1}{k}$

By the Hockey-Stick Identity, $\displaystyle\sum^{m}_{i=1}\dbinom{k+i-1}{k}=\dbinom{k+m}{k+1}=\dbinom{(k+1)+m-1}{k+1}$ , and we have completed our proof.

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It remains to find the number of zeroes to the right of the last non-zero digit in $f(2014,2014)=\binom{4027}{2014}=\dfrac{4027!}{2013!2014!}$ . We can do this by counting the number of zeroes in $4027!$ , then subtracting the number of zeroes in $2013!$ and $2014!$

The number of zeroes in $4027!$ is $\displaystyle\sum_{i=1}^{\infty}\left\lfloor\dfrac{4027}{5^i}\right\rfloor=1005$ .

The number of zeroes in $2013!$ is $\displaystyle\sum_{i=1}^{\infty}\left\lfloor\dfrac{2013}{5^i}\right\rfloor=501$ .

The number of zeroes in $2014!$ is $\displaystyle\sum_{i=1}^{\infty}\left\lfloor\dfrac{2014}{5^i}\right\rfloor=501$ .

Thus the number of zeroes after the last non-zero digit of $\dbinom{4027}{2014}$ is, perhaps surprisingly, $1005-501-501=\boxed{3}$ .