Toán 6 nâng cao?

H

hocvuima

hoanganh530 said:
Cho B=3−32+33−34+...+31999−32000.3-3^{2}+3^{3}-3^{4}+...+3^{1999}-3^{2000}.
Khi đó ta có: 3-4B=3n3^{n}. n=?
Bấm để xem đầy đủ nội dung ...

Ta có:
B=3−32+33−34+...+31999−32000B=3-3^{2}+3^{3}-3^{4}+...+3^{1999}-3^{2000}
B.3=32−33+34−35+...+32000−32001B.3=3^2-3^{3}+3^{4}-3^{5}+...+3^{2000}-3^{2001}
B.3=32−33+34−35+...+32000−32001−3+3B.3=3^2-3^{3}+3^{4}-3^{5}+...+3^{2000}-3^{2001}-3+3
B.3=3−32001−BB.3=3-3^{2001}-B
B.4=3−32001B.4=3-3^{2001}
3−B.4=32001=3n3-B.4=3^{2001}=3^{n}
Vậy n=2001n=2001

Đúng nhưng không có chỗ xác nhận sorry
 
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