For what natural numbers n in N,is the inequa | vedclass

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(B) Given the inequality: $2^n > n+1$. For $n=1$: $2^1 > 1+1 \Rightarrow 2 > 2$,which is false. For $n=2$: $2^2 > 2+1 \Rightarrow 4 > 3$,which is true

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$\forall n \geq 2$

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(B) Given the inequality: $2^n > n+1$. For $n=1$: $2^1 > 1+1 \Rightarrow 2 > 2$,which is false. For $n=2$: $2^2 > 2+1 \Rightarrow 4 > 3$,which is true. For $n=3$: $2^3 > 3+1 \Rightarrow 8 > 4$,which is true. Using the principle of mathematical induction,assume $2^k > k+1$ for some $k \geq 2$. We want to show $2^{k+1} > (k+1)+1 = k+2$. Since $2^k > k+1$,multiplying both sides by $2$ gives $2^{k+1} > 2k+2$. We know $2k+2 = (k+2) + k$. Since $k \geq 2$,$k > 0$,therefore $2k+2 > k+2$. Thus,$2^{k+1} > k+2$ is true for all $n \geq 2$.

For what natural numbers $n \in N$,is the inequality $2^n > n+1$ valid?

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