Question

Calculati urmatoarele sume si apoi verificati rezultatul obtinut folosind metoda inductiei matematice:


1/1*2 + 1/2*3 + ... + 1/n(n+1).

Answer 1

$\frac{1}{1*2}=\frac{1}{1}-\frac{1}{2}\\\\1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}\\\\p(n):\ \frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{n*(n+1)}=1-\frac{1}{n+1},\ n\ apartine\ lui\ N^{*}\\\\Verificare:\\\\p(1):\ \frac{1}{1*2}=1-\frac{1}{1+1}\ =>\ \frac{1}{2}=\frac{1}{2}\ Adevarat\\\\Demonstratie:\\\\p(k)=>p(k+1),\ =>\ inseamna\ implica$

$p(k):\ \frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{k*(k+1)}=1-\frac{1}{k+1},\ k\ apartine\ lui\ N^{*},\ Adevarat\\\\p(k+1):\ \frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{k*(k+1)}+\frac{1}{(k+1)*(k+2)}=1-\frac{1}{k+2},\ k\ apartine\ lui\ N^{*}\\\\p(k+1):\ p(k)+\frac{1}{(k+1)*(k+2)}=1-\frac{1}{k+2}\\\\p(k+1):\ 1-\frac{1}{k+1}+\frac{1}{(k+1)*(k+2)}=1-\frac{1}{k+2}\\\\-\frac{1}{k+1}+\frac{1}{(k+1)*(k+2)}=-\frac{1}{k+2}\\\\\frac{1}{(k+1)*(k+2)}=\frac{1}{k+1}-\frac{1}{k+2}$

$\frac{1}{(k+1)*(k+2)}=\frac{k+2}{(k+1)*(k+2)}-\frac{k+1}{(k+1)*(k+2)}\\\\\frac{1}{(k+1)*(k+2)}=\frac{k+2-k-1}{(k+1)*(k+2)}\\\\\frac{1}{(k+1)*(k+2)}=\frac{1}{(k+1)*(k+2)}\ Adevarat\\\\Atunci\ p(n)\ Adevarata\ oricare\ ar\ fi\ n\ apartine\ lui\ N^{*}$