Tổng S=2019C0 + 2019C3 + 2019C6 + ... + 2019C2019 bằng:

11/15

Tổng S=C20190+C20193+C20196+...+C20192019 bằng:

22019-23.

22019+43.

22019+23.

22019-43.

Giải thích

Đáp án cần chọn là: A

Ta tìm các số phức z thỏa mãn z3=1 ta có:

z3=1⇔z3-1=0↔z-1z2+z+1=0⇔z=1z2+z+1=0⇔z1=1z2=-12+32iz3=-12-32i

Xét kai triển

1+x2019=∑k=02019C2019kxk=C20190+C20191x+C20192x2+...+C20192019x2019(*)

Thay z2=-12+32i vào khai triển (*) ta được

1-12+32i2019=C20190+C20191z2+C20192.z22+...+C20192019z22019⇔12+32i2019=C20190+C20191z2+C20192z22+C20193+...+C20192019⇔-1=C20190+C20193+C20196+...+C20192019+z2C20191+C20194+...+C20192017 +z22C20192+C20195+...+C20192018 (1)

Tương tự thay z3=-12-32i vào khai triển (*) ta được:

-1-12-32i2019=C20190+C20191z3+C20192.z32+...+C20192019z32019⇔12-32i2019=C20190+C20191z3+C20192z32+C20193+...+C20192019⇔-1=C20190+C20193+C20196+...+C20192019+z3C20191+C20194+...+C20192017 +z32C20192+C20195+...+C20192018 (2)

Thay z=1 vào khai triển (*) ta được:

22019=C20190+C20191+C20192+...+C20192019⇔22019=C20190+C20193+C20196+...+C20192019+C20191+C20194+C20197+...+C20192017 +C20192+C20195+C20198+...+C20192018 (3)

Cộng vế với vế của (1); (2); (3) ta được:

22019-2=3C20190+C20193+...+C20192019+1+z2+z3C20191+C20194+...+C20192018 +1+z22+z32C20192+C20195+...+C20192017

Nhận thấy 1+z2+z3=1-12+32i-12-32i=0 và 1+z22+z32=1+-12+32i2+-12-32i2=0

Nên 22019-2=3S⇔S=22019-23.