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15 tháng 6 2019

9 tháng 7 2018

C = 1+1/2(1+2)+1/3(1+2+3)+........+1/2015(1+2+3+4+...+2015) 

C = 1 + \(\frac{1}{2}\cdot\frac{2.3}{2}\)\(\frac{1}{3}\cdot\frac{3.4}{2}\)+ ... + \(\frac{1}{2015}\cdot\frac{2015.2016}{2}\)

C = \(\frac{2}{2}\) + \(\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\)

C = \(\frac{2+3+4+...+2016}{2}\)

Đặt D = 2 + 3 + 4 + ... + 2016 

Số số hạng của D là : (2016 - 2) : 1 + 1 = 2015

Tổng D là :  (2 + 2016) . 2015 : 2 = 2033135

Thay D vào biểu thức C ta được : \(\frac{2033135}{2}\)

Vậy C = ... . 

6 tháng 6 2015

a/A=1+2+4+8+...+1024

2A=2+4+8+16+....+2048

2A-A=(2+4+8+16+....+2048)-(1+2+4+8+...+1024)

A=2048-1

A=2047

VẬY A=2047

b/B=1+5+25+125+....+15625

5B=5+25+125+625+....+78125

5B-B=(5+25+125+625+....+78125)-(1+5+25+125+....+15625)

4B=78125-1

4B=78124

B=78124:4

B=19531

VẬY B =19531

C=1/1.2+1/2.3+1/3.4+...+1/2015.2016

C=1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016

=1-1/2016

=2015/2016

VẬY C=2015/2016

D/=10/1.3+10/3.5+10/5.7+....+10/2013.2015

=5(2/1.3+2/3.5+2/5.7+...+2/2013.2015)

=5(1-1/3+1/3-1/5+1/5-1/7+..+1/2013-1/2015)

=5(1-1/2015)

=5.2014/2015

=2014/403

VẬY D=2014/403

 

6 tháng 6 2015

a, A = 1 + 2 + 4 + 8 +...+ 1024

   \(A=1+2+2^2+2^3+....+2^{10}\)

   \(2A=2+2^2+2^3+....+2^{10}+2^{11}\)

   \(A=1+2+2^2+2^3+....+2^{10}\)

  \(A=2^{11}-1=2047\)

b, B = 1 + 5 + 25 + 125 + ... + 15625

   \(B=1+5+5^2+5^3+....+5^6\)

   \(3B=5+5^2+5^3+....+5^6+5^7\)

   \(B=1+5+5^2+5^3+....+5^6\)

  \(2B=5^7-1\Rightarrow B=\frac{5^7-1}{2}=39062\)

d, D = 10 / 1 . 3 + 10 / 3 . 5 + 10 / 5 . 7 + ... + 10 / 2013 . 2015

\(D=\frac{10}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(D=\frac{10}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(D=\frac{10}{2}.\left(1-\frac{1}{2015}\right)=5.\frac{2014}{2015}=\frac{2014}{403}\)

Câu c thì tương tự

9 tháng 5 2019

\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)

\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)

Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)

\(=2017\)

Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)

\(=2017\)