Question

ajutorla inductiile matematice,P(n)=1/1.3+1/3.5+...+1/(2n-1)(2n+1)=n/2n+1

Answer 1

$P(n): \frac{1}{1*3}+ \frac{1}{3*5} + ... + \frac{1}{(2n-1)(2n+1)}= \frac{n}{2n+1}, ~n \geq 1 \\ \\ I)~Verificarea: \\ \\ P(1): \frac{1}{1*3} = \frac{1}{2*1+1} ~~~-\ \textgreater \ ~~~ \frac{1}{3}= \frac{1}{3}~(adevarat) \\ \\ \\ II) Demonstratia: \\ \\ P(k)~-\ \textgreater \ ~P(k+1) \\ \\ P(k): \frac{1}{1*3}+ \frac{1}{3*5}+...+ \frac{1}{(2k-1)(2k+1)}= \frac{k}{2k+1}~(o~presupunem~ca~fiind~ \\ adevarata)$
$P(k+1): \frac{1}{1*3}+ \frac{1}{3*5}+...+ \frac{1}{(2k-1)(2k+1)}+ \frac{1}{[2(k+1)-1][2(k+1)+1]}= \\ \\ = \frac{k+1}{2(k+1)+1}~(de~la~\frac{1}{1*3}~pana~la~ \frac{1}{(2k-1)(2k+1)}~este~chiar~P(k)~deci~putem~ \\ \\ inlocui~astfel) \\ \\ P(k+1): \frac{k}{2k+1} + \frac{1}{(2k+2-1)(2k+2+1)}= \frac{k+1}{2k+2+1} \\ \\ P(k+1): \frac{k}{2k+1}+ \frac{1}{(2k+1)(2k+3)}= \frac{k+1}{2k+3}~(numitorul~comun~e~ \\ \\ (2k+1)(2k+3) ) \\ \\ P(k+1):k(2k+3)+1=(k+1)(2k+1) \\ \\ P(k+1): 2k^{2}+3k+1= 2k^{2}+k+2k+1$
$P(k+1):2k^{2}+3k+1= 2k^{2}+3k+1~(adevarat) \\ \\ Din~I)~si~II) ~-\ \textgreater \ ~ \\ \\ -\ \textgreater \ ~P(n): \frac{1}{1*3}+ \frac{1}{3*5} +...+ \frac{1}{(2n-1)(2n+1)}= \frac{n}{2n+1}~adevarat~ \\ \\ pentru~orice~n \geq 1. \\ \\ \\ Sper~ca~te-am~ajutat!$