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Fibonacci Number(f_(n-1)+f_n=f_(n+1)) , f_0=0 , f_1=1, prove that f_n=f_(k+1)*f_(n-k)+f_k*f(n-k-1)
I want to use induction to prove that , so I need to assume when n=k is true , then move on to prove k+1 is true , but in this case , I need to assume k and k+1 are true , but can I assume k and k+1 are true , it's a little wierd assuming too much .
Numbers are the essence of the Universe
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I want to use induction to prove that , so I need to assume when n=k is true , then move on to prove k+1 is true , but in this case , I need to assume k and k+1 are true, but can I assume k and k+1 are true , it's a little wierd assuming too much .
No, if you want to prove k+1 is true, you can't assume k+1 is true. What you can do however is assume that k and all integers less than k are true. This is known as "strong" induction, and I believe it's what you want to use.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Doesn't matter, you can assume k and k+1, but you have to prove k+2 case from them. And you need to verify both k=0 and k=1.
X'(y-Xβ)=0
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