prove that (n_C_0)^2+(n_C_1)^2+...+(n_C_n)^2=2n_C_n
(one more thing in case of Kalk, no induction :D)
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prove that (n_C_0)^2+(n_C_1)^2+...+(n_C_n)^2=2n_C_n
(one more thing in case of Kalk, no induction :D)
Lol...why no induction, bugz?:D
because it can be done without induction!
(x+1)^(2n)=(x+1)^n * (x+1)*n
the LHS has a term of 2n_C_n *x^n
the RHS has x^n with coefficients (n_C_0)(n_C_n)+(n_C_1)(n_C_n-1)+...+(n_C_n)(n_C_1)
so (n_C_0)=(n_C_n) and (n_C_1)=(n_C_n-1) and ...
so therefore
2n_C_n=Sum (r=0..n) n_C_r
Ok bugz, I'm getting sick of your elegant solutions! You know way too much for the rest of us!:D
Hi, just a trivial matter, but it may be easier to understand if you use HTML tags. i.e. instead of n_C_r, wriet nCr
Oh, and I take it your last line was meant to say:
(2n)_C_n=Sum (r=0..n) n_C_r ^2
But anyway, its a nice proof.