is it valid in this type of proof to place any value of the k < 6 on one of the side to setup our equations)# for n=k e.g :
\( (k+1)!> 2^k .2 -1 +2^k -k \)
can we take k<4 * or *k=2on the R.H.S to proof our question

@Syed Hissaan
–
No, because the statement is not true for \( k \leq 4 \). That is why you have to start the base case in the appropriate domain of \( n > 5 \) (or whatever they tell you to).

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TopNewest@Calvin Lin @Chung Kevin , any proof for that one !

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What have you tried? What have you done?

Seems to me like a simple induction proof starting with base case of \( n = 6 \).

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is it valid in this type of proof to place any value of the k < 6 on one of the side to setup our equations)# for n=k e.g : \( (k+1)!> 2^k .2 -1 +2^k -k \) can we take

k<4on the R.H.S to proof our question* or *k=2Log in to reply

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