Tính tổng T= 0 C2020/ 3- 1C2020/4+ 2C2020/5- 3C 2020/6 +....-2019C2020/2022

Chọn C

Ta có (1−x)2020=∑k=02020C2020k(−x)k 

Xét hàm số:

f(x)=x2(1−x)2020=∑k=02020C2020kxk+2.(−1)k=C20200x2−C20201x3+C20202x4−C20203x5+...+C20202019x2021−C20200x2022 

Là hàm số liên tục trên ℝ nên:

∫01x2(1−x)2020dx=∫01(C20200x2−C20201x3+C20202x4−C20203x5+...+C20202019x2021−C20200x2022)dx 

Ta xét

VT=∫01x2(1−x)2020dx   (đặt u=1−x⇒x=1−u⇒dx=−du;   đổi biến x=0⇒u=1;    x=1⇒u=0) 

=∫10(1−u)2u2020(−du)=∫01(u2020−2u2021+u2022)du=u20212021−2u20222022+u2023202301=12021−22022+12023=14133456313 

VP=∫01(C20200x2−C20201x3+C20202x4−C20203x5+...−C20202019x2021+C20200x2022)dx=C20200.x33−C20201.x44+C20202.x55−C20203.x66+...−C20202019.x20222022+C20202020.x2023202301=C202003−C202014+C202025−C202036+....−C202020192022+C202020202023. 

VT=VP⇒C202003−C202014+C202025−C202036+....−C202020192022+C202020202023=14133456313.