¿Demostración de sumatorias por inducción?

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Hola Diego Perez!

Esa igualdad está mal escrita, te sobra el 1 inicial

$$\begin{align}&\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2n-1)(2n+1)}= \frac n {2n+1}\\&\\&Para\ n=1\\&\frac 1 3= \frac 1 3\\&\\&Para \ n=2 \\&\\&\frac 1 {1·3}+ \frac 1 {3 ·5} \stackrel{?}{=}\frac{2} 5\\&\\&\frac{5+1}{15}= \frac 6 {15}= \frac 2 5\\&\\&Supongamos \ se \ cumple \ para \ n=k\\&\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2k-1)(2k+1)}= \frac k {2k+1}\\&\\&¿Se \ cumplirá  \ para\ n=k+1?\\&\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2k-1)(2k+1)}+\frac 1 {2(k+1)(k+3)}\stackrel{?}{=} \frac{ k+1} {2k+3}\\&\\&\Bigg[\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2k-1)(2k+1)}\Bigg]+\frac 1 {(2k+1)(k+3)}=\\&\\&\Bigg[\frac k {2k+1} \Bigg ]+\frac 1 {(2k+1)(k+3)}=\\&\\&\frac {k(2k+3)+1}{(2k+1)(2k+3)}= \frac {2k^2+3k+1}{(2k+1)(2k+3)}=\\&\\&factorizandoº el \ numeradorº (buscando \ sus \ raíces)\\&\\&=\frac {2(k+\frac 1 2)(k+1)}{(2k+1)(2k+3)}=\frac {(2k+ 1 )(k+1)}{(2k+1)(2k+3)}=\frac {k+1}{2k+3}\\&c.q.d.\end{align}$$

como queríamos demostrar

Saludos

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Copie algunas líneas mal:

$$\begin{align}&\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2n-1)(2n+1)}= \frac n {2n+1}\\&\\&Para\ n=1\\&\frac 1 3= \frac 1 3\\&\\&Para \ n=2 \\&\\&\frac 1 {1·3}+ \frac 1 {3 ·5} \stackrel{?}{=}\frac{2} 5\\&\\&\frac{5+1}{15}= \frac 6 {15}= \frac 2 5\\&\\&Supongamos \ se \ cumple \ para \ n=k\\&\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2k-1)(2k+1)}= \frac k {2k+1}\\&\\&¿Se \ cumplirá  \ para\ n=k+1?\\&\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2k-1)(2k+1)}+\frac 1 {[2(k+1)-1][2(k+1)+1]}\stackrel{?}{=} \frac{ k+1} {2k+3}\\&\\&\Bigg[\frac 1 {1·3}+ \frac 1 {3 ·5} + \frac 1 {5·7}+····+ \frac 1 {(2k-1)(2k+1)}\Bigg]+\frac 1 {(2k+1)(2k+3)}=\\&\\&\Bigg[\frac k {2k+1} \Bigg ]+\frac 1 {(2k+1)(2k+3)}=\\&\\&\frac {k(2k+3)+1}{(2k+1)(2k+3)}= \frac {2k^2+3k+1}{(2k+1)(2k+3)}=\\&\\&factorizandoº el \ numeradorº (buscando \ sus \ raíces)\\&\\&=\frac {2(k+\frac 1 2)(k+1)}{(2k+1)(2k+3)}=\frac {(2k+ 1 )(k+1)}{(2k+1)(2k+3)}=\frac {k+1}{2k+3}\\&c.q.d.\end{align}$$

ahora si

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