proof that 2^(n+1) >(n+2)sin n


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Question Number 162305 by mathlove last updated on 28/Dec/21

$p r o o f t h a t$ $2^{n + 1} > (n + 2) \sin n$
	Answered by aleks041103 last updated on 28/Dec/21

	$s i n (n) ⩽ 1$ $\Rightarrow (n + 2) s i n (n) ⩽ n + 2$ $f o r n = 1 :$ $2^{n + 1} = 4 > 3 = n + 2$ $f o r n = k :$ $2^{k + 1} > k + 2$ $f o r n = k + 1 :$ $2^{(k + 1) + 1} = 22^{k + 1} > 2 (k + 2) = 2 k + 4$ $k ⩾ 1 \Rightarrow 2 k + 4 ⩾ k + 1 + 4 = k + 5 > (k + 1) + 2$ $\Rightarrow 2^{(k + 1) + 1} > (k + 1) + 2$ $\Rightarrow f o r \forall n \in N, 2^{n + 1} > n + 2 ⩾ (n + 2) s i n (n)$ $\Rightarrow n = 1, 2, . . ., 2^{n + 1} > (n + 2) s i n (n)$ $f o r n = 0 :$ $2^{n + 1} = 2 > 0 = (n + 2) s i n (n)$ $\Rightarrow 2^{n + 1} > (n + 2) s i n (n), \forall n \in N^{0}$
	Answered by mr W last updated on 28/Dec/21

	$r e c a l l :$ $1 ⩾ \sin n f o r a n y n \in R$ ${(1 + a)}^{n} > 1 + n a f o r a > 0$ $2^{n + 1} = {(1 + 1)}^{n + 1}$ $> 1 + (n + 1) \times 1 = n + 2 = (n + 2) \times 1$ $⩾ (n + 2) \sin n$ $\Rightarrow 2^{n + 1} > (n + 2) \sin n ✓$
	Commented byaleks041103 last updated on 28/Dec/21

	$I t i s w o r t h n o t i n g t h a t$ $(n + 2) ⩾ (n + 2) s i n (n)$ $i s t r u e f o r n ⩾ - 2$
	Commented bymr W last updated on 29/Dec/21

	$i a s s u m e d n \in N w h a t t h e q u e s t i o n e r$ $a l s o m e a n t, i t h i n k .$ $i f n \in Z, t h e n 2^{n + 1} > (n + 2) \sin n i s$ $t r u e f o r n ⩾ - 6 .$
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