Estude! Com Prof. Caju

calcule o valor da soma [tex3]a_{1},a_{2},a_{3},...,a_{n}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]a_{1},a_{2},a_{3},...,a_{n}[/tex3]

da sequencia [tex3]\frac{1}{3},\frac{1}{15},\frac{1}{35},...,\frac{1}{(2n-1)(2n+1)}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{3},\frac{1}{15},\frac{1}{35},...,\frac{1}{(2n-1)(2n+1)}[/tex3]

[tex3]S=\sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\sum_{k=1}^{n}\frac{(2k+1)-(2k-1)}{(2k-1)(2k+1)}=\frac{1}{2}(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k+1})⇒2S=(1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1})-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n+1})⇒2S=1-\frac{1}{2n+1}⇒S=\frac{n}{2n+1}.[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]S=\sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\sum_{k=1}^{n}\frac{(2k+1)-(2k-1)}{(2k-1)(2k+1)}=\frac{1}{2}(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k+1})⇒2S=(1+\frac{1}{3}+\frac{1}{5}+\dots+\frac{1}{2n-1})-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n+1})⇒2S=1-\frac{1}{2n+1}⇒S=\frac{n}{2n+1}.[/tex3]

Olá, NigrumCibum, porque o 2S ?

raquelcds, eu multipliquei por 2 dos dois lados pra sumir com o 1/2 que estava na frente do somatório.