Bài tập Chuyên đề Nhị thức Newton có đáp án

Tính: a) S=C2022092022+C2022192021+…+C2022k92022−k+…+C202220219+C20222022. b) T=C2022042022−C2022142021 . 3+…−C202220214 . 32021+C2022202232022.

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Câu hỏi: Tính: a) S=C2022092022+C2022192021+...+C2022k92022−k+...+C202220219+C20222022. b) T=C2022042022−C2022142021 . 3+...−C202220214 . 32021+C2022202232022. Trả lời: Áp dụng công thức nhị thức Newton ta có: a) S=C2022092022+C2022192021+...+C2022k92022−k+...+C202220219+C20222022 =C2022092022+C2022192021 . 1+...+C2022k92022−k.1k+...+C202220219.12021+C20222022.12022 =9+12020=102022.…

Chứng minh: Cn03n+Cn13n−1+…+Cnk3n−k+…+Cnn−13+Cnn=Cn0+Cn13+…+Cnk3k+…+Cnn−13n−1+Cnn3n

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Câu hỏi: Chứng minh: Cn03n+Cn13n−1+...+Cnk3n−k+...+Cnn−13+Cnn=Cn0+Cn13+...+Cnk3k+...+Cnn−13n−1+Cnn3n Trả lời: Cn03n+Cn13n−1+...+Cnk3n−k+...+Cnn−13+Cnn =Cn03n+Cn13n−1.11+...+Cnk3n−k.1k+...+Cnn−13.1n−1+Cnn.1n =3+1n=4n. Cn0+Cn13+...+Cnk3k+...+Cnn−13n−1+Cnn3n =Cn0.1n+Cn1.1n−13+...+Cnk.1n−k3k+...+Cnn−1.1n−13n−1+Cnn3n =1+3n=4n. Vậy Cn03n+Cn13n−1+...+Cnk3n−k+...+Cnn−13+Cnn =Cn0+Cn13+...+Cnk3k+...+Cnn−13n−1+Cnn3n. ====== **** mời các bạn…

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