Mother of Summations!

Using repeatedly the Hockey Stick identity : $\Large\displaystyle\sum_{\color{#D61F06}{z}=1}^{2016}\sum_{\color{#D61F06}{y}=1}^z\cdots\sum_{\color{#D61F06}{c}=1}^d\sum_{\color{#D61F06}{b}=1}^c\sum_{\color{#D61F06}{a}=1}^b(1)$ $\Large =\displaystyle\sum_{\color{#D61F06}{z}=1}^{2016}\sum_{\color{#D61F06}{y}=1}^z\cdots\sum_{\color{#D61F06}{c}=1}^d\sum_{\color{#D61F06}{b}=1}^c\binom b1$ $\Large =\displaystyle\sum_{\color{#D61F06}{z}=1}^{2016}\sum_{\color{#D61F06}{y}=1}^z\cdots\sum_{\color{#D61F06}{c}=1}^d\binom{c+1}{2}$ $\cdots\\\cdots$ $\Large =\displaystyle\sum_{\color{#D61F06}{z}=1}^{2016}\sum_{\color{#D61F06}{y}=1}^z\binom{y+23}{24}$ $\Large =\displaystyle\sum_{\color{#D61F06}{z}=1}^{2016}\binom{z+24}{25}=\binom{2041}{26}$ $\Large\therefore 80(\color{#69047E}{26})-\color{#0C6AC7}{2041}+10=49$ $\Huge \therefore \sqrt{49}=\color{#20A900}{\boxed 7}$

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If you're looking for a proof here are some cute proofs posted... :-)

I have a slightly different combinatorial way of solving the problem.

The above summation actually represents the number of ways we can form a sequence of 26 natural numbers: $z,y,x...b,a$ such that each number is less than or equal to the previous number (when scanned from left) and $z<2017$ .

For solving this, consider $2041$ identical boxes arranged in a row and $26$ balls which have to be put in these boxes. The number of empty boxes to the left of $k^{th}$ ball $+1$ will represent the number at the $k^{th}$ position in the sequence. It is easy to see that there is a bijection between the two.

Therefore the final answer is simply the number of ways of putting $26$ balls in $2041$ boxes which is:

$\dbinom{2041}{26}$