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# Inequality proof

Prove that if $$n>2$$, then $$2^{n+1} > n^2 + n + 2$$ holds true.

Note by Abdelfatah Teamah
1 year, 1 month ago

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Proof By Induction:

Base Case: $$n=3$$

$$2^{3+1}>3²+3+2\implies 16>14$$,which is true.

Inductive Step: Let the statement be true for some $$k$$.We will prove that it holds for $$k+1$$ as well.

By the inductive hypothesis we have $$2^{k+1}>k²+k+2$$

We have to prove that $$2^{k+2}>(k+1)²+(k+1)+2$$

Proof: $2^{k+2}=2(2^{k+1})>2(k²+k+2)>(k+1)²+(k+1)+2$ The last inequality, after simplifying becomes $$k²>k$$,which is true as $$k>2$$.

Hence Proved.

- 1 year, 1 month ago