Question

Tính 

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

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)

Answer 1

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

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

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

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

Answer 2

giúp câu P luôn với bạn

Answer 3

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

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

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

Answer 4

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

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

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\right)\)

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

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

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

\(P=\frac{\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}}\)

Đặt \(F=\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)

Tách 2004 thành tổng của 2004 số 1

\(\Rightarrow F=\frac{2005}{2005}+\left(\frac{2003}{2}+1\right)+\left(\frac{2002}{3}+1\right)+...+\left(\frac{1}{2004}+1\right)\)

\(F=\frac{2005}{2005}+\frac{2005}{2}+\frac{2005}{3}+...+\frac{2005}{2004}\)

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

Thay F vào P ta có:

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

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

Vậy \(P=\frac{1}{2005}\)

Answer 5

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

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

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+.......+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+......+\frac{1}{3^{2015}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2015}}\)

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

Vậy ....

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.........+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+........+\frac{1}{2004}}\)

Xét tử số:

\(\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+........+\frac{1}{2004}\)

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

\(=2005.\left(\frac{1}{2}+\frac{1}{3}+...........+\frac{1}{2005}\right)\)

Thay vào ta có:

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

Vậy .....