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

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

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

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

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

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

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

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

\(=2017\)

minh nghi ban dang chep sai de bai 

the de bai cua minh thi giai nhu sau

A=1/2015.2016+1/2016.2017+......+1/2029.2030

A=1/2015-1/2016+1/2016-1/2017+.....+1/2029-1/2030

A=1/2015-1/2030=3/818090

a)\(=\frac{2017}{2016}.\frac{3}{4}-\frac{1}{2016}.\frac{3}{4}\)

\(=\frac{3}{4}\left(\frac{2017}{2016}-\frac{1}{2016}\right)\)

\(=\frac{3}{4}.1\)

\(=\frac{3}{4}\)

b)\(=\frac{2015}{2016}\left(\frac{1}{2}+\frac{1}{3}-\frac{5}{6}\right)\)

\(=\frac{2015}{2016}.0\)

\(=0\)

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

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

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

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

\(=2017\)

a)

số số hạng của phép tính a) là:

(2016-1):1+1=1016(số)

tổng của phép tính trên là:

(2016+1)x1016:2=1024636

b)

số số hạng của phép tính b) là:

(2017-3):2+1=1008(số)

tổng của phép tính trên là:

(2017+3)x1008:2=1018080

kb z mk nhé cac bn

a,Số số hạng trong dãy là:(2016-1):1+1=2016

   Tổng của dãy là:(1+2016).2016:2=2 033 136

b,Số số hạng của dãy là:(2017-3):2+1=1008

   Tổng của dãy là:(3+2017).1008:2=1 018 080

Số số hạng là 

(2016-1):1+1=2016(số)

Tổng là 

(2016+1)x2016:2=2033136

   Đáp số ....

Số số hạng là 

(2017-3):2+1=1008

Tổng là 

(2017+3)x1008:2=1018080

   Đáp số