soluţii şi barem

Solutie: a1= 1, a2= 1, a3= 8, a4= 3, a5= 21 si se arata prin inductie ca:Într-adevar, daca a = n− 1 atunci:an⎧n−1, npar= ⎨, ( ∀)n≥4.2⎩n − n+1, n imparn3 3⎡ n ⎤ ⎡n−1 an+11 ⎤ ⎡ 2n n1 ⎤ 2n n n n2( ) ( )= ⎢ = + = + + 1+ = + + 1= + 1 − + 1 + 1n−1 ⎥ ⎢n−1 n−1 ⎥ ⎢⎣ n−1⎥.⎣ ⎦ ⎣ ⎦⎦⎡ ( n+ 1) 3 ⎤ ⎡( n+ 1)( n 2 + n+ 1)+ n 2 + n⎤⎡ n 2 + n ⎤an+2= ⎢ ⎥= ⎢⎥ = n+ 1+ = n+12 2 2n + n+ 1 n + n+ 1⎢n + n+1⎥ .⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦⎣ ⎦Atunci a2006= 2005 si lima2n=+∞= lim a2n+ 1deci lim a n=+∞ .n→∞n→∞2 2⎧ n ⎡ n ⎤n→∞⎪ − , n par2 2 2n 1⎢n 1⎥⎧n ⎫ n ⎡n⎤ ⎪ − ⎣ − ⎦Fie bn= ⎨ ⎬= − ⎢ ⎥= ⎨=2 2⎩an ⎭ an ⎣an⎦ ⎪ n ⎡ n ⎤⎪− , n impar2 2n − n+ 1 ⎢n − n+1 ⎥⎩ ⎣ ⎦2⎧ n2⎡ 1n 1⎤, n par⎧ n⎧ 1− + + − ( n+1, ) n par , n par⎪n−1 ⎢⎣ n−1⎥ ⎦ ⎪n 1 ⎪=−n−1⎨ =2⎨ =2⎨. Atunci⎪ n ⎡ n−1⎤nn−1− 1 + , n impar ⎪ −1, n impar ⎪ , n impar2 2 22⎪n n 1 ⎢ n n 1⎥ ⎩ − + ⎣ − + ⎦ ⎪⎩n − n+1⎪⎩n − n+1limb = limb = 0⇒ limb= 0.2n 2n+1nn→∞ n→∞ n→∞