Tính tổng sau đây: C2021^0 - 2.C2021^1 + 2^2.C2021^2 - 2^3.C2021^3 + ... - 2^2021.C2021^2021

Tính tổng sau đây:

$C_{2021}^0 - 2.C_{2021}^1 + {2^2}.C_{2021}^2 - {2^3}.C_{2021}^3 + ... - {2^{2021}}.C_{2021}^{2021}$.

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 $C_{2021}^0 - 2.C_{2021}^1 + {2^2}.C_{2021}^2 - {2^3}.C_{2021}^3 + ... - {2^{2021}}.C_{2021}^{2021} = - 1$.