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4 tháng 8 2018

\(D=\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2014}}\)

\(D=\frac{2.2014}{\frac{2}{2}+\frac{1}{\frac{2.3}{2}}+...+\frac{1}{\frac{2015.2014}{2}}}\)

\(D=\frac{2.2014}{\frac{2}{2}+\frac{2}{2.3}+...+\frac{2}{2014.2015}}\)

\(D=\frac{2015}{\frac{1}{2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}}\)

\(D=\frac{2014}{\frac{1}{2}+\frac{1}{2}-\frac{1}{2015}}\)

4 tháng 8 2018

\(D=\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}}\)

\(D=\frac{2.2014}{\frac{1}{\frac{\left(1+1\right).1}{2}}+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2014+1\right).2014}{2}}}\)

\(D=\frac{2.2014}{\frac{2}{1.2}+\frac{2}{3.2}+\frac{2}{4.3}+\frac{2}{2015.2014}}\)

\(D=\frac{2.2014}{2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)}\)

\(D=\frac{2014}{\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)}\)

\(D=\frac{2014}{\left(1-\frac{1}{2015}\right)}\)

\(D=\frac{2014}{\frac{2014}{2015}}\)

\(D=\frac{2014.2015}{2014}\)

\(D=2015\)

Tham khảo nhé~

\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)

\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)

\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)

\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)

\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)

\(A=2.2014.\left(1-\frac{1}{2014}\right)\)

\(A=2.2014.\frac{2013}{2014}\)

\(A=\frac{2.2014.2013}{2014}\)

\(A=2.2013\)

\(A=4026\)

4 tháng 1 2017

A=4026

DD
27 tháng 5 2021

\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)

\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)

\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)

\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)

\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)

\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)

\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)

19 tháng 8 2017

Đặt phân thức trên là D

=> D=(1+1+1+1+...+1+2013/2+2012/3+...+2/2013+1/2014)/(1/2+1/3+1/4+...+1/2014)

=> D=(1+2013/2+1+2012/3+1+2011/4+...+1+2/2013+1+1/2014+1)/(1/2+1/3+1/4+1/5+...+1/2014)

=> D=(2015/2+2015/3+2015/4+...+2015/2013+2015/2014+1)/(1/2+1/3+1/4+...+1/2014)

=> D=[2015*(1/2+1/3+1/4+1/5+....+1/2014)]/(1/2+1/3+1/4+1/5+...+1/2014)

=> D=2015

13 tháng 6 2020

UwU

ư uwsuuuuuuuuuuuu kimochiiiiiiiiiiiiiiiiiiii

đùa thôi đáp án: 2015 nha bn

ư ư wsuuuuuuuuuuuuuuuuuuuuuuuuuu kimmmmmooooochiiiiiiiiiii

À quên nhớ FOLOW CHO TUI NHA!

24 tháng 7 2017

Ta có: \(\frac{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...\frac{1}{2014}+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=

\(\frac{\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)+1+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=

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

AH
Akai Haruma
Giáo viên
22 tháng 10

Lời giải:

$M=1+\frac{1}{2}.\frac{2(2+1)}{2}+\frac{1}{3}.\frac{3(3+1)}{2}+\frac{1}{4}.\frac{4(4+1)}{2}+....+\frac{1}{2014}.\frac{2014(2014+1)}{2}$
$=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2015}{2}$

$=\frac{2+3+4+....+2015}{2}$

$=\frac{1+2+3+....+2015}{2}-\frac{1}{2}$
$=\frac{2015(2015+1)}{4}-\frac{1}{2}=\frac{2031119}{2}$

16 tháng 3 2016

nhớ phải 4 k thì làm

17 tháng 3 2016

tớ cần gấp !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!