Question: Let $F_m$ be the mth Fibonacci number given by $F_1=F_2=1$ and $F_{m+2} = F_m + F_{m+1}$ for all $m\ge1$.
Show that $\sum_m(C_{n,k})=F_{m+1}$ ; here the sum is over all pairs of integers $n\ge k\ge0$ with $n+k=m$.
My Approach: $F_3=2,F_4=3,F_5=5$;
Now for $m=4$: $F_5=5$ and $(n,k)$ for $m=4$ are $(4,0),(3,1),(2,2)$, so $\sum_{n+k=4} C_{n,k}=C_{4,0} + C_{3,1} + C_{2,2} = 1+3+1 = 5 $.
Next how can i proceed??
We have that $$ \eqalign{ & S(m) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m - k \cr k \cr} \right)} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m - 1 - k \cr k \cr} \right) + \left( \matrix{ m - 1 - k \cr k - 1 \cr} \right)} = \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m - 1} \right)} {\left( \matrix{ m - 1 - k \cr k \cr} \right)} + \sum\limits_{\left( {1\, \le } \right)\,k\,\left( { \le \,m - 1} \right)} {\left( \matrix{ m - 2 - \left( {k - 1} \right) \cr k - 1 \cr} \right)} = \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m - 1} \right)} {\left( \matrix{ m - 1 - k \cr k \cr} \right)} + \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m - 2} \right)} {\left( \matrix{ m - 2 - j \cr j \cr} \right)} = \cr & = S(m - 1) + S(m - 2) \cr} $$ which is the same recurrence as for Fibonacci Numbers, however $$ \left\{ \matrix{ S(0) = 1 \hfill \cr S(1) = 1 \hfill \cr} \right. $$ and therefore $$ S(m) = F_{\,m + 1} \quad \left| {\;0 \le m} \right. $$
