Explicație pas cu pas:
Avem:
[tex]P(n): 7+77+777+...+77..77 (de~n~ori)=\frac{7*(10^{n+1}-9*n-10)}{81}, n~natural~\geq 1[/tex]
Aplicam principiul intai al inductiei matematice.
1) Etapa de verificare:
Daca n=1, atunci:
[tex] P(1): 7=\frac{7*(10^{1+1}-9*1-10)}{81}=\frac{7*(100-9-10)}{81}=\frac{7*81}{81}=7[/tex]
Deci, P(1) este adevarata.
2) Etapa inductiva:
Presupunem P(k) adevarata, unde:
[tex]P(k): 7+77+777+...+77..77 (de~k~ori)=\frac{7*(10^{k+1}-9*k-10)}{81}[/tex]
Demonstram P(k+1) adevarata, unde:
[tex]P(k+1): 7+77+777+...+77..77 (de~k+1~ori)=\frac{7*(10^{k+1+1}-9*(k+1)-10)}{81}[/tex]
Observam ca:
[tex]7+77+777+...+77..77 (de~k+1~ori)-(7+77+777+...+77..77 (de~k~ori))=7+77+777+...+77..77(de~k~ori)+77..77 (de~k+1~ori)-7-77-777-...-77..77(de~k~ori)=77..77(de~k+1~ori)[/tex]
Aratam aceasta egalitate, folosindu-ne de datele prezentate mai sus. Facem scaderea relatiilor matematice prezente in P(k+1) si P(k).
[tex]\frac{7*(10^{k+1+1}-9*(k+1)-10)}{81}-\frac{7*(10^{k+1}-9*k-10)}{81}=\frac{7*(10^{k+1+1}-9*(k+1)-10)-(7*(10^{k+1}-9*k-10))}{81}=\frac{7*(10^{k+2}-9k-9-10)-7*10^{k+1}+7*9k+7*10}{81}=\frac{7*10^{k+2}-7*10^{k+1}-7*9k+7*9k-7*9-70+70}{81}=\frac{7*10^{k+1}(10-1)-63}{81}=\frac{7*9*10^{k+1}-63}{81}=\frac{63*(10^{k+1}-1)}{81}=\frac{7*(10^{k+1}-1)}{9}[/tex]
[tex]\frac{7*(10^{k+1}-1)}{9}=\frac{7*(10*10*...*10(de~k+1~ori)-1)}{9}=\frac{7*99..99(de~k+1~ori)}{9}=7*11...11(de~k+1~ori)=77...77(de~k+1~ori)[/tex]
Numarul obtinut este exact forma unui numar reprezentat prin cifre de 7.
Asadar, si P(k+1) este adevarata.
Deci, din 1) si 2) conform principiului I al inductiei matematice P(n) este adevarata.
