IME / ITA ⇒ (Farias Brito) Sequências Tópico resolvido

[Gu178]

Prove que:[tex3]1-\frac{1}{2}+\frac{1}{3} - \frac{1}{4}+...+\frac{1}{2013} - \frac{1}{2014}=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2013}+\frac{1}{2014}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]1-\frac{1}{2}+\frac{1}{3} - \frac{1}{4}+...+\frac{1}{2013} - \frac{1}{2014}=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2013}+\frac{1}{2014}[/tex3]

[Auto Excluído (ID:12031)]

[tex3]\sum_{n=1}^{2014} \frac{(-1)^{n+1}}n = \sum_{n=1}^{1007} \frac1{n+1007}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=1}^{2014} \frac{(-1)^{n+1}}n = \sum_{n=1}^{1007} \frac1{n+1007}[/tex3]

[tex3]\sum_{n=1}^{2014} \frac{(-1)^{n+1}}n = \sum_{n=1}^{2014} \frac1{n} - \sum_{n=1}^{1007}\frac1n[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=1}^{2014} \frac{(-1)^{n+1}}n = \sum_{n=1}^{2014} \frac1{n} - \sum_{n=1}^{1007}\frac1n[/tex3]

[tex3]\sum_{n=1}^{1007}\frac1{2n-1}-\sum_{n=1}^{1007} \frac{1}{2n} = \sum_{n=1}^{2014} \frac1{n} - \sum_{n=1}^{1007}\frac1n[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=1}^{1007}\frac1{2n-1}-\sum_{n=1}^{1007} \frac{1}{2n} = \sum_{n=1}^{2014} \frac1{n} - \sum_{n=1}^{1007}\frac1n[/tex3]

[tex3]\sum_{n=1}^{1007}(\frac1{2n-1}- \frac{1}{2n}) = \sum_{n=1}^{2014} \frac1{n} - \sum_{n=1}^{1007}\frac1n[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=1}^{1007}(\frac1{2n-1}- \frac{1}{2n}) = \sum_{n=1}^{2014} \frac1{n} - \sum_{n=1}^{1007}\frac1n[/tex3]

[tex3]\sum_{n=1}^{1007}(\frac1{2n-1}- \frac{1}{2n} + \frac1n) = \sum_{n=1}^{2014} \frac1{n}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=1}^{1007}(\frac1{2n-1}- \frac{1}{2n} + \frac1n) = \sum_{n=1}^{2014} \frac1{n}[/tex3]

[tex3]\sum_{n=1}^{1007}(\frac1{2n-1}+ \frac{1}{2n}) = \sum_{n=1}^{2014} \frac1{n}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sum_{n=1}^{1007}(\frac1{2n-1}+ \frac{1}{2n}) = \sum_{n=1}^{2014} \frac1{n}[/tex3]

o que é verdade

[MateusQqMD]

Outra forma

[tex3]1-\frac{1}{2}+\frac{1}{3} - \frac{1}{4}+...+\frac{1}{2013} - \frac{1}{2014} = 1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4}+...+\frac{1}{2013} + \frac{1}{2014} - 2\Big[\frac{1}{2}+\frac{1}{4} + \frac{1}{6}+ \ .\ .\ .\ + \frac{1}{2014}\Big][/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]1-\frac{1}{2}+\frac{1}{3} - \frac{1}{4}+...+\frac{1}{2013} - \frac{1}{2014} = 1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4}+...+\frac{1}{2013} + \frac{1}{2014} - 2\Big[\frac{1}{2}+\frac{1}{4} + \frac{1}{6}+ \ .\ .\ .\ + \frac{1}{2014}\Big][/tex3]

[tex3]= 1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4} \ +\ .\ .\ .\ +\ \frac{1}{2013} + \frac{1}{2014} \ \ - \Big[1+\frac{1}{2} + \frac{1}{3}+ \ .\ .\ . \ + \frac{1}{1007}\Big][/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]= 1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4} \ +\ .\ .\ .\ +\ \frac{1}{2013} + \frac{1}{2014} \ \ - \Big[1+\frac{1}{2} + \frac{1}{3}+ \ .\ .\ . \ + \frac{1}{1007}\Big][/tex3]

[tex3]= \frac{1}{1008}+\frac{1}{1009}+ \ .\ .\ .\ +\ \frac{1}{2013}+\frac{1}{2014}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]= \frac{1}{1008}+\frac{1}{1009}+ \ .\ .\ .\ +\ \frac{1}{2013}+\frac{1}{2014}[/tex3]