=>3A= 3^2017-3^2016+3^2015-...-3^2+3

=>3A+A=4A=3^2017+1=>A=\(\frac{3^{2017}+1}{4}\)

B tương tự nha

\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)

\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)

\(B=\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+,...+\left(-2014\right)+2016\)

\(B=2+2+....+2\left(\text{504 số hạng 2}\right)=1008\)

c) 1 + ( -3 ) +5 + ( -7 ) + ...........+ 2013 + ( -2015 )

[ 1 + (-3 ) ] + [ 5 + -7 ] + .......... + [ 2013 + ( - 2015 ) ]

 có số cặp là : [ ( 2015 - 1 ) : 2 + 1 ] : 2 = 504 ( cặp )

= -2 + -2 + -2 +..........+ -2

= -2 x 504

= -1008

Ta có:

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

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

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

chị kết bạn với em nha gửi lời kết bn với em nhé

j zậy em hả 

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, s1 có 2015 hạng tử

=> s1= (2014:2).-1+2015=1007.(-1)+2015=1008

Lời giải:

a,S1=1+(-2)+3+(-4)+...+(-2014)+2015

=(1-2)+(3-4)+...+(2013-2014)+2015

=-1+(-1)+...+(-1)+2015

=-1.1007+2015

=(-1007)+2015

=1008

b,S2=(-2)+4+(-6)+8+...+(-2014)+2016

=(-2+4)+(-6+8)+...+(-2014+2016)

=2+2+...+2

=2.504

=1008

c,S3=1+(-3)+5+(-7)+...+2013+(-2015)

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

=(-2)+(-2)+...+(-2)

=(-2).504

=-1008

d,S4=(-2015)+(-2014)+(-2013)+...+2015+2016

=(-2015+2015)+...+0+2016

=0+...+0+2016

=2016

STUDY WELL !

Cô ơi dấu hiệu chia hết cho 5 em mở không được