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

Note by Abdelfatah Teamah 6 months, 2 weeks 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. – Abdur Rehman Zahid · 6 months, 1 week ago

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TopNewestProof 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. – Abdur Rehman Zahid · 6 months, 1 week ago

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