Cho B= \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\) . Chứng minh B<\(\frac{1}{2}\)
cho B =\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)chứng minh rằng B < \(\frac{1}{2}\)
\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)
\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)
3B=1+1/3+1/32+...+1/32004
3B-B=1-1/32005
2B=1-1/32005
B=1/2-1/(32005.2)
Vậy B <1/2
Hùng ơi sai rồi
3B=1+1/3+1/3^2+...+1/3^2004 chứ
Thay số 3 thành 1 vì 1/3*3=1 ko phải bằng 3
Bài 2 :
a ) Tính A = \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right)\div\frac{1890}{2005}+115\)
b ) Cho :
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
Chứng minh :
\(B< \frac{1}{2}\)
\(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
Ta nhận thấy các cặp số đều bằng 3/5 và các dấu cũng giống nhau. ( các số có cùng dấu thì phân số đó cũng cùng dấu.)
=> Phân số này sẽ bằng 3/5
\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\)
Ta nhận thấy các cặp số đều bằng -3/5 và các dấu thì trái nhau. ( các số có trái dấu thì phân số đó cũng trái dấu.)
=> Phân số này sẽ bằng -3/5.
Sau khi rút gọn bài toán sẽ thành:
\(\left(\frac{3}{5}-\frac{3}{5}\right)\div\frac{1890}{2005}+115=115\)
Câu b tạm thời mình chưa nghĩ ra. Chúc bạn học tốt.
a) \(A=\left(\frac{3}{5}-\frac{3}{5}\right):\frac{1890}{2005}+115\)
\(\Rightarrow A=115\)
b) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+....+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
\(\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)
\(\Rightarrow B< \frac{1}{2}\)
Tính
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)
\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)
b) Đặt \(A=\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)
\(\Rightarrow A=\left(\frac{2003}{2}+1\right)+\left(\frac{2002}{3}+1\right)+...+\left(\frac{1}{2004}+1\right)+1\) ( tách 2004/1=2004 ra, cộng cho các phân số kia mỗi phân số 1 đơn vị, thì còn dư ra 1)
\(A=\frac{2005}{2}+\frac{2005}{3}+...+\frac{2005}{2004}+\frac{2005}{2005}\) ( 1 = 2005/2005)
\(A=2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)\)
Thay A vào P được
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)}\)
\(P=\frac{1}{2005}\)
Tính:
B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+..+\frac{1}{2004}}\)
Bài 2 : Tính : B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
Mẫu số = 2004/1 + 2003/2 + 2002/3 + ... + 1/2004
= (1 + 1 + ... + 1) + 2003/2 + 2002/3 + ... + 1/2004
2004 số 1
= (1 + 2003/2) + (1 + 2002/3) + ... + (1 + 1/2004) + 1
= 2005/2 + 2005/3 + ... + 2005/2004 + 2005/2005
= 2005 × (1/2 + 1/3 + ... + 1/2004 + 1/2005)
=> B = 1/2005
Mẫu số = 2004/1 + 2003/2 + 2002/3 + ... + 1/2004
= (1 + 1 + ... + 1) + 2003/2 + 2002/3 + ... + 1/2004
2004 số 1
= (1 + 2003/2) + (1 + 2002/3) + ... + (1 + 1/2004) + 1
= 2005/2 + 2005/3 + ... + 2005/2004 + 2005/2005
= 2005 × (1/2 + 1/3 + ... + 1/2004 + 1/2005)
=> B = 1/2005
Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
P=\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}\)-\(\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)
Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Cho B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
CMR:B<\(\frac{1}{2}\)