Prove-that-2-n-2-gt-n-2-for-n-N-by-mathematical-induction-

Question Number 126124 by ZiYangLee last updated on 17/Dec/20

Prove that 2^{n} + 2 > n^{2} for n \in N by

mathematical induction .

Commented by talminator2856791 last updated on 17/Dec/20

i was having the time of my life till i read “ by mathematical {induction}^{″}!

hahahhhhahha

Commented by ZiYangLee last updated on 30/Dec/20

Good luck haha

Answered by mahdipoor last updated on 17/Dec/20

w e k n o w : n ⩾ 4 \Rightarrow n^{2} ⩾ 2 n + 1 (*)

\forall n ⩾ 4 \Rightarrow i f 2^{n} ⩾ n^{2} \Rightarrow 2 \times 2^{n} ⩾ 2 \times n^{2} \Rightarrow

2^{n + 1} ⩾ n^{2} + n^{2} \overset{*}{\Rightarrow} 2^{n + 1} ⩾ n^{2} + 2 n + 1 \Rightarrow

2^{n + 1} ⩾ {(n + 1)}^{2} (* *)

f o r n = 4 \Rightarrow 2^{n} ⩾ n^{2}

\overset{* *}{\Rightarrow} f o r : n = 4 + 1 = 5 \Rightarrow 2^{n} ⩾ n^{2}

\overset{* *}{\Rightarrow} f o r : n = 5 + 1 = 6 \Rightarrow 2^{n} ⩾ n^{2}

\dots .

\Rightarrow f o r : \forall n ⩾ 4 \Rightarrow 2^{n} ⩾ n^{2} \Rightarrow 2^{n} + 2 > n^{2}

f o r : n = 1, 2, 3 \Rightarrow 2^{n} + 2 > n^{2}

\Rightarrow\Rightarrow f o r : \forall n \in N \Rightarrow 2^{n} + 2 > n^{2}

Answered by physicstutes last updated on 17/Dec/20

prove for n = 1, {(2)}^{1} + 2 > 1^{2}

assume for n = k

\Rightarrow 2^{k} + 2 > k^{2}

prove for n = k + 1 .

2^{k + 1} + 2 = 2^{k} . 2 + 2 = 2 (2^{k} + 1)

from 2^{k} + 2 > k^{2}

\Rightarrow 2 (2^{k - 1} + 1) > k^{2}

but \forall n \in N, 2^{k - 1} < 2^{k}

\Rightarrow 2 (2^{k} + 1) > {(k + 1)}^{2}

thus true \forall n \in N