Determinant-

\\\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}

4 Answers

341
Hari Shankar ·

Well \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

1
Che ·

thanks

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

341
Hari Shankar ·

I 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

1
Che ·

ok

but one thing

wen k>n/2

n-k < k

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