a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)

\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)

\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)

giúp câu P luôn với bạn

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

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)}\)

\(P=\frac{1}{2005}\)

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

Ta có:

\(\frac{1\div2003+1\div2004-1\div2005}{5\div2003+5\div2004-5\div2005}\)    -     \(\frac{2\div2002+2\div2003-2\div2004}{3\div2002+3\div2003-3\div2004}\)

Đơn giản đi hết ta sẽ còn:

\(\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)

2.

Ta có: 

Số khoảng cách của các số trong dãy là  2³ = 8

=> Tổng của dãy dưới sẽ gấp 8 lần tổng dãy trên.

=> 3025 . 8 = 24200

\(\Rightarrow3B=3+\frac{1}{3^1}+\frac{1}{3^2}+....+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=\left(3+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=3-\frac{1}{3^{2005}}\Rightarrow B=\left(3-\frac{1}{3^{2005}}\right):2\)

\(\Rightarrow\left(3-\frac{1}{3^{2005}}\right):2<\frac{1}{2}\Rightarrow B<\frac{1}{2}\)

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

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)

\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)

\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)

Ta có:

\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)

\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)

\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)

Có : 

3B = 1 + 1/3 + 1/3^2 + .... + 1/3^2004

2B = 3B - B = ( 1 + 1/3 + 1/3^2 + ..... + 1/3^2004 ) - ( 1/3 + 1/3^2 + 1/3^3 + ..... + 1/3^2005 )

                  = 1 - 1/3^2005 < 1

=> B < 1 : 2 = 1/2

=> ĐPCM

Tk mk nha

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)

\(\Rightarrow B< \frac{1}{2}\)

Kết bạn LQM ko

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