Binomial identity with a parameter L.

36 Views Asked by At

This identity arose during my attempts to disprove the existence of integer loops in the Collatz Conjecture. Is there a name for it? It contains a selectable parameter, $L \geq 1.$ I remember seeing the factor (n-k+1) somewhere.

$L\geq 1, \ n\geq 1.$

$$ \sum_{k=1}^{n}(n-k+1) \binom{L+k-2}{L-1}= \binom{L+n}{L+1}$$

Example 1.

L=1, n=6. Then, $$ \sum_{k=1}^{n=6}(n-k+1) \binom{L+k-2}{L-1}= \binom{L+n}{L+1}=\sum_{k=1}^{n=6}(7-k)\binom{k-1}{0}=6\binom{0}{0}+5\binom{1}{0}+4\binom{2}{0}+3\binom{3}{0}+2\binom{4}{0}+1\binom{5}{0}$$ $$ =6*1 + 5*1 + 4*1 + 3*1 + 2*1 + 1*1 = \binom{7}{2}=21$$

Example 2.

L=4, n=5. We have then,

$$ \sum_{k=1}^{n=5}(n-k+1) \binom{L+k-2}{L-1}= \binom{L+n}{L+1}=\sum_{k=1}^{n=5}(6-k)\binom{k+2}{3}=5\binom{3}{3}+4\binom{4}{3}+3\binom{5}{3}+2\binom{6}{3}+1\binom{7}{3}=$$ $$ =5*1 + 4*4 + 3*10 + 2*20 + 1*35=\binom{9}{5}=126$$