Giải bài 2 trang 37

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: ${(9 + x)}^{2022} = C_{2022}^{0} 9^{2022} . x^{0} + C_{2022}^{1} 9^{2021} . x^{1} + . . . + C_{2022}^{k} 9^{2022 - k} . x^{k} + . . . + C_{2022}^{2021} 9. x^{2021} + C_{2022}^{2022} . x^{2022}$

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

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

${(4 + x)}^{2022} = C_{2022}^{0} 4^{2022} . x^{0} + C_{2022}^{1} 4^{2021} . x^{1} + . . . + C_{2022}^{k} 4^{2022 - k} . x^{k} + . . . + C_{2022}^{2021} 4. x^{2021} + C_{2022}^{2022} . x^{2022}$

Thay $x = - 3$ ta được

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