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6 tháng 8 2016

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

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)

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

6 tháng 8 2016

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

\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(\Leftrightarrow3A-A=1-\frac{1}{3^{2005}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{2005}}}{2}\)

11 tháng 10 2020

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

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

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

11 tháng 10 2020

Ta có:

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

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

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

23 tháng 12 2016

Bài 1:

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

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

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

\(\Rightarrow P=\frac{-7}{15}\)

Vậy \(P=\frac{-7}{15}\)

Bài 2:
Ta có: \(S=23+43+63+...+203\)

\(\Rightarrow S=13+10+20+23+...+103+100\)

\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)

\(\Rightarrow S=3025+450\)

\(\Rightarrow S=3475\)

Vậy S = 3475

23 tháng 12 2016

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

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

P = \(\frac{3}{15}-\frac{10}{15}\)

=> P =\(\frac{-7}{15}\)

2. ta có:

S = 23 + 43 + 63 +...+ 203

=> S = 13 + 10 + 23 + 20 +...+ 103 + 100

=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )

=> S = 3025 + 550

=> S = 3575

Vậy S = 3575

12 tháng 7 2017

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

11 tháng 7 2018

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

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

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

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

11 tháng 7 2018

giúp câu P luôn với bạn

20 tháng 7 2016

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

20 tháng 7 2016

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

19 tháng 3 2020

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

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

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

5 tháng 8 2018

\(\forall n\inℕ^∗\)ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)

\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)  (*)

Thay n=1; n=2; n=3; .....; n=2004 Ta có:

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\)

\(=1-\frac{1}{\sqrt{2005}}\)