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Question Number 106816 by bemath last updated on 07/Aug/20

$prove by mathematical induction$ $(1) n^{2} ⩽ 2^{n}; n ⩾ 4$ $(2) {(n + 1)}^{2} < {2 n}^{2}; n ⩾ 3$ $(3) 2^{n} - 3 ⩾ 2^{n - 2}; n ⩾ 5$ $@ bemath @$
Commented by bobhans last updated on 07/Aug/20

$(2) n^{2} + 2 n + 1 < {2 n}^{2}; 2 n + 1 < n^{2}; n ⩾ 3$ $P_{1} (n = 3) \Rightarrow 2.3 + 1 < 3^{2} [true]$ $let : P_{k} (n = k, k ⩾ 3) \Rightarrow 2 k + 1 < k^{2} [true]$ $P_{k + 1} (n = k + 1) \Rightarrow LHS : 2 (k + 1) + 1 = 2 k + 3$ $= (2 k + 1) + 2 < k^{2} + 2 = k^{2} - 2 k + 1 + (1 + 2 k)$ $= {(k + 1)}^{2} + 2 k + 1 =$
	Answered by bobhans last updated on 07/Aug/20

	$(3) 2^{n} - 3 ⩾ \frac{2^{n}}{4} \Rightarrow 4 . 2^{n} - 12 ⩾ 2^{n}$ $3 . 2^{n} ⩾ 12 \Rightarrow 2^{n} ⩾ 4 \Rightarrow n ⩾ 2 .$ $wrong to condition n ⩾ 5 .$
	Answered by Rio Michael last updated on 07/Aug/20

	$(1) claim : n^{2} ⩽ 2^{n}; if n ⩾ 4$ $prove for n = 4 \Rightarrow 4^{2} = 2^{4} = 16$ $prove for n = 6 \Rightarrow 6^{2} ⩽ 2^{6}$ $Assume that n = k satisfies the claim .$ $∴ k^{2} ⩽ 2^{k} : if k ⩾ 4$ $proving for n = k + 1$ $\Rightarrow {(k + 1)}^{2} ⩽ 2^{k + 1}$ $\Rightarrow k^{2} + 2 k + 1 ⩽ 2 (2^{k})$ $but k^{2} ⩽ 2^{k} \Rightarrow k^{2} ⩽ 2 (2^{k})$ $also k^{2} ⩾ 2 k \forall k \in Z$ $thus k^{2} + 2 k + 1 ⩽ 2 (2^{k}) if k ⩾ 3$
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