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can anyone show me how to solve this? possile solution

Prove by induction a result of the form “for all n ≥ T, 2n < n!”. Use the best possible value of T.

studymaths
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2 Answers2

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Well you first need to conjecture some value of $T$. You see that the inequality is wrong for $T=1,2,3$ but true for $T=4$. So here is our conjecture : $$ n \geq 4 \implies 2n < n! $$ Then you prove it by induction. The base case will indeed be at $n=4$ and is trivial.

Regarding the induction step you can say : $$ (n+1)! = n!\cdot (n+1) \overset{(*)}{>} 2n(n+1) $$ $(*)$ inductive hypothesis.

I let you finish.

Zubzub
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This can be easily said. We take $T=4$.

Then $$n!=n×(n-1)×... ×2×1$$. Thus $n!=2n×((n-1)!÷2)$.

(n-1)!÷2 is always greater than 1 for all n > 3. We do not need induction the problem is obvious.

Suprabha
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