\\\texttt{If} \: f_{n}=\sum_{k=0}^{n}\binom{n-k}{k} \\\texttt{evaluate}\\ \\ \begin{vmatrix} f_{n}& f_{n+1}&f_{n+2} \\ f_{n+1}& f_{n+2}&f_{n+3}\\ f_{n+2}&f_{n+3}& f_{n+4} \end{vmatrix}
341Well \sum_{k=0}^n \binom{n-k}{k} is the expression for the nth Fibonacci number.
So fn is the nth Fibonacci number.
And we have the recursion f_{m+2} = f_{m+1} + f_m
So by R_3 \rightarrow R_2+R_3 you can make two rows identical and so the determinant is zero
1thanks
btw
this \sum_{k=0}^{n}\binom{n-k}{k} means nth Fibonacci number.........din knew that
i was thinking it to be (n-k)Ck :P
341I think you have misunderstood. That isnt a representation.
\sum_{k=0}^n ^{n-k}C_k = f_n where fn is the nth Fibonacci number