Ensino Superior ⇒ Álgebra Elementar

[Nekololikuro]

Prove, por indução finita, que vale a desigualdade:
[tex3]\frac{1}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{2}[/tex3]

.[tex3]\frac{3}{4}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{3}{4}[/tex3]

.[tex3]\frac{5}{6}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{5}{6}[/tex3]

...[tex3]\frac{2n-1}{2n}\leq \frac{1}{\sqrt[]{3n+1}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{2n-1}{2n}\leq \frac{1}{\sqrt[]{3n+1}}[/tex3]

, [tex3]\forall [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\forall [/tex3]

n [tex3]\in \mathbb{N}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\in \mathbb{N}[/tex3]

[PeterPark]

[tex3]\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}~~~~~~~~~~\forall n,~n\geq1 \\\ \\\ \\\ \\\ \\\ \bf{P(1).~~ n=1} \\\ \\\ \frac{2-1}{2}\leq \frac{1}{\sqrt{4}}~~~~~~\rightarrow ~~~~~~\frac{1}{2}\leq \frac{1}{2} ~~~~~~~~~~~~~~~~~~~~OK\\\ \\\ \\\ \\\ \\\ P(x).~~n=x \\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2x-1}{2x}\leq \frac{1}{\sqrt{3x+1}}~~~~~~~~~~~~~~~~~~~H\\\ \\\ \\\ \\\ \\\ P(x+1).~~n=x+1 \\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3(x+1)+1}} \\\ \\\ \\\ \\\ P(x)\rightarrow P(x+1).[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}~~~~~~~~~~\forall n,~n\geq1 \\\ \\\ \\\ \\\ \\\ \bf{P(1).~~ n=1} \\\ \\\ \frac{2-1}{2}\leq \frac{1}{\sqrt{4}}~~~~~~\rightarrow ~~~~~~\frac{1}{2}\leq \frac{1}{2} ~~~~~~~~~~~~~~~~~~~~OK\\\ \\\ \\\ \\\ \\\ P(x).~~n=x \\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2x-1}{2x}\leq \frac{1}{\sqrt{3x+1}}~~~~~~~~~~~~~~~~~~~H\\\ \\\ \\\ \\\ \\\ P(x+1).~~n=x+1 \\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3(x+1)+1}} \\\ \\\ \\\ \\\ P(x)\rightarrow P(x+1).[/tex3]

Na hipótese:
[tex3]\bf\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2x-1}{2x}\cdot \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3x+1}}\cdot \frac{2(x+1)-1}{2(x+1)} \\\ \\\ \\\ Provando~~que: \\\ \\\ \frac{1}{\sqrt{3x+1}}\cdot \frac{2(x+1)-1}{2(x+1)}\leq\frac{1}{\sqrt{3(x+1)+1}}\rightarrow \\\ \\\ \\\ \sqrt{3(x+1)+1}\cdot(2x+1) \leq\sqrt{3x+1}\cdot (2x+2) \\\ \\\ (3x+4)\cdot(4x^2+4x+1)\leq(3x+1)\cdot (4x^2+8x+4) \\\ \\\ 12x^3+28x^2+19x+4 \leq 12x^3+28x^2+20x+4 \\\ \\\ x\geq0 ~~~~~~~~~~~~~~OK \\\ \\\\ \\\ \\\ \\\ Então~~na~~hipótese:\\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2x-1}{2x}\cdot \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3x+1}}\cdot \frac{2(x+1)-1}{2(x+1)}\leq\frac{1}{\sqrt{3(x+1)+1}}~~~~~~~~~~~~~~~~~P(x)~~~~~\rightarrow \\\ \\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3(x+1)+1}}~~~~~~~~~~~~~~~~~~~~~~~P(x+1)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\bf\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2x-1}{2x}\cdot \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3x+1}}\cdot \frac{2(x+1)-1}{2(x+1)} \\\ \\\ \\\ Provando~~que: \\\ \\\ \frac{1}{\sqrt{3x+1}}\cdot \frac{2(x+1)-1}{2(x+1)}\leq\frac{1}{\sqrt{3(x+1)+1}}\rightarrow \\\ \\\ \\\ \sqrt{3(x+1)+1}\cdot(2x+1) \leq\sqrt{3x+1}\cdot (2x+2) \\\ \\\ (3x+4)\cdot(4x^2+4x+1)\leq(3x+1)\cdot (4x^2+8x+4) \\\ \\\ 12x^3+28x^2+19x+4 \leq 12x^3+28x^2+20x+4 \\\ \\\ x\geq0 ~~~~~~~~~~~~~~OK \\\ \\\\ \\\ \\\ \\\ Então~~na~~hipótese:\\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2x-1}{2x}\cdot \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3x+1}}\cdot \frac{2(x+1)-1}{2(x+1)}\leq\frac{1}{\sqrt{3(x+1)+1}}~~~~~~~~~~~~~~~~~P(x)~~~~~\rightarrow \\\ \\\ \\\ \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\dots \frac{2(x+1)-1}{2(x+1)}\leq \frac{1}{\sqrt{3(x+1)+1}}~~~~~~~~~~~~~~~~~~~~~~~P(x+1)[/tex3]

[tex3]P(1)\rightarrow P(2)\rightarrow P(3)......\rightarrow P(x)\rightarrow P(x+1)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]P(1)\rightarrow P(2)\rightarrow P(3)......\rightarrow P(x)\rightarrow P(x+1)[/tex3]