Ta có: Với(x2 + 1) n ax12 (1 + x2) n =n∑ k=0 Cknx 2n = C 0n + C 1nx2+. . . +Cknx 12−2k x = 1 : 2 n = C0 n + C1 n+. . . +Cn n = 1024 ⇔ 2 n = 1024 ⇔ n = 10
(x2 + 1) n ax 12 (1 + x 2) n = n∑ k=0 C knx 2n = C 0n + C 1nx 2+. . . +C knx 12−2k x = 1 : 2 n = C 0 n + C 1 n+. . . +C n n = 1024 ⇔ 2 n = 1024 ⇔ n = 10