# Giải Bài 2 Trang 37 Chuyên đề Học Tập Toán 10 – Cánh Diều>

Đề bài

Tính:

a) $$S = C_{2022}^0{9^{2022}} + C_{2022}^1{9^{2021}} + … + C_{2022}^k{9^{2022 – k}} + … + C_{2022}^{2021}9 + C_{2022}^{2022}$$

b) $$T = C_{2022}^0{4^{2022}} – C_{2022}^1{4^{2021}}.3 + … – C_{2022}^{2021}{4.3^{2021}} + C_{2022}^{2022}{.3^{2022}}$$

Phương pháp giải – Xem chi tiết

Công thức nhị thức Newton: $${(a + b)^n} = C_n^0{a^n} + C_n^1{a^{n – 1}}b + … + C_n^{n – 1}a{b^{n – 1}} + C_n^n{b^n}$$

Lời giải chi tiết

a) Theo công thức nhị thức Newton, ta có: $${\left( {9 + x} \right)^{2022}} = C_{2022}^0{9^{2022}}.{x^0} + C_{2022}^1{9^{2021}}.{x^1} + … + C_{2022}^k{9^{2022 – k}}.{x^k} + … + C_{2022}^{2021}9.{x^{2021}} + C_{2022}^{2022}.{x^{2022}}$$

Thay $$x = 1$$ ta được: $${\left( {9 + 1} \right)^{2022}} = S = C_{2022}^0{9^{2022}} + C_{2022}^1{9^{2021}} + … + C_{2022}^k{9^{2022 – k}} + … + C_{2022}^{2021}9 + C_{2022}^{2022} \Rightarrow S = {10^{2022}}$$

b) Theo công thức nhị thức Newton, ta có:

$${\left( {4 + x} \right)^{2022}} = C_{2022}^0{4^{2022}}.{x^0} + C_{2022}^1{4^{2021}}.{x^1} + … + C_{2022}^k{4^{2022 – k}}.{x^k} + … + C_{2022}^{2021}4.{x^{2021}} + C_{2022}^{2022}.{x^{2022}}$$

Thay $$x = – 3$$ ta được

$$\begin{array}{l}{\left( {4 – 3} \right)^{2022}} = C_{2022}^0{4^{2022}}.{\left( { – 3} \right)^0} + C_{2022}^1{4^{2021}}.{\left( { – 3} \right)^1} + …… + C_{2022}^{2021}4.{\left( { – 3} \right)^{2021}} + C_{2022}^{2022}.{\left( { – 3} \right)^{2022}}\\ \Leftrightarrow {1^{2022}} = T = C_{2022}^0{4^{2022}} – C_{2022}^1{4^{2021}}.3 + … – C_{2022}^{2021}{4.3^{2021}} + C_{2022}^{2022}{.3^{2022}}\\ \Leftrightarrow T = 1\end{array}$$