\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)

\(S=1-\frac{1}{2016}=\frac{2015}{2016}\)

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

\(S=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+......+\left(-\frac{1}{2015}+\frac{1}{2015}\right)-\frac{1}{2016}\)

\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

1,-2015

2,50

3,-2015

thiện xạ 5a3 có thể giải chi tiết ra đc k? Mk cần cách lm

2017/2016

S = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/2014x2015 + 1/2015x2016

S = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... +  1/2014 - 1/2015 + 1/2015 - 1/2016

S = 1 - 1/2016

S = 2015

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

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

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

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

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 !

bằng 15 hay sao ý