Tính tổng sau đây: C2021^0 - 2.C2021^1 + 2^2.C2021^2 - 2^3.C2021^3 + ... - 2^2021.C2021^2021
Giải thích
Lời giải
Ta có \(C_{2021}^0 - 2.C_{2021}^1 + {2^2}.C_{2021}^2 - {2^3}.C_{2021}^3 + ... - {2^{2021}}.C_{2021}^{2021}\).
\( = C_{2021}^0 + C_{2021}^1.\left( { - 2} \right) + C_{2021}^2.{\left( { - 2} \right)^2} + C_{2021}^3.{\left( { - 2} \right)^3} + ... + C_{2021}^{2021}.{\left( { - 2} \right)^{2021}}\).
\[ = C_{2021}^0{.1^{2021}} + C_{2021}^1{.1^{2020}}.\left( { - 2} \right) + C_{2021}^2{.1^{2019}}.{\left( { - 2} \right)^2} + C_{2021}^3{.1^{2018}}.{\left( { - 2} \right)^3} + ... + C_{2021}^{2021}.{\left( { - 2} \right)^{2021}}\].
= [(1 + (–2)]2021 = (–1)2021 = –1.
Vậy \(C_{2021}^0 - 2.C_{2021}^1 + {2^2}.C_{2021}^2 - {2^3}.C_{2021}^3 + ... - {2^{2021}}.C_{2021}^{2021} = - 1\).