Hãy tìm hai số hạng đứng giữa các khai triển:
a) $ ( a^3 + ab)^{31} $
b) $ ( a^3 + ba)^{30} $ 
a.    Khai triển của nhị thức  $ {\left( {{a^3} + ab} \right)^{31}} $ có 32 số hạng.
Do đó hai số hạng giữa là số hạng thứ 16 và 17
Ta có:     $ {T_{16}} = C_{31}^{15}{\left( {{a^3}} \right)^{16}}{\left( {ab} \right)^{15}} = C_{31}^{15}{a^{63}}{b^{15}} $ ;  $ {T_{17}} = C_{31}^{16}{\left( {{a^3}} \right)^{15}}{\left( {ab} \right)^{16}} = C_{31}^{15}{a^{61}}{b^{16}} $
b.    Khai triển nhị thức  $ {\left( {{a^3} + ba} \right)^{30}} $ có 31 số hạng nên chỉ có 1 số hạng giữa là 16
Ta có:     $ {T_{16}} = C_{31}^{15}{\left( {{a^3}} \right)^{14}}{\left( {ab} \right)^{15}} = C_{31}^{15}{a^{57}}{b^{15}} $

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