số số hạng của mẫu dãu số :

   `(2018 - 2) : 1 + 1= 2017(số hạng)`

tổng mãu dãu trên là :

    ` (2018 + 2) xx 2017 : 2 = 2037170`

=> `1/2 + 1/3 + 1/4 + ... + 1/2018 = 1/2037170`

a, S = 1 + 2 + 2² + 2³ + ... + 2²⁰¹⁷

Ta có : 2S = 2 + 2² + 2³ +.... + 2²⁰¹⁸

Lấy 2S - S ta được : S = 2²⁰¹⁸ - 1

b, Đặt S = 3 + 3² + 3³ + ... + 3²⁰¹⁷

Ta có : 3S = 3² + 3³ + ... + 3²⁰¹⁸

Lấy 3S - S ta được 2S = 3²⁰¹⁸ -3 

=> \(S=\frac{3^{2018}-3}{2}\)

c, Đặt S = 4 + 4² + 4³ + ... + 4²⁰¹⁷ 

Ta có : 4S = 4² + 4³ + ... + 4²⁰¹⁸

Lấy 4S - S ta được 3S = 4²⁰¹⁸ - 4 

=> \(S=\frac{4^{2018}-4}{3}\)

a, S = 1 + 2 + 22 + 23 + ... + 22017

Ta có : 2S = 2 + 22 + 23 +.... + 22018

Lấy 2S - S ta được : S = 22018 - 1

b, Đặt S = 3 + 32 + 33 + ... + 32017

Ta có : 3S = 32 + 33 + ... + 32018

Lấy 3S - S ta được 2S = 32018 -3 

=> �=32018−32S=232018−3​

c, Đặt S = 4 + 42 + 43 + ... + 42017 

Ta có : 4S = 42 + 43 + ... + 42018

Lấy 4S - S ta được 3S = 42018 - 4 

=> �=42018−43S=342018−4​
 

Đặt A=1^2-2^2+......-2016^2+2017^2

-A=2^2-1^2+........+2016^2-2015^2

ÁP dụng A^2-B^2=(A+B)(A-B)

sau đó bạn sẽ tính được A

Đặt \(S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)

 Biến đổi mẫu 

\(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)

\(=\left(2017+1\right)+\left(\frac{2016}{2}+1\right)+...+\left(\frac{1}{2017}+1\right)-2017\)

\(=2018+\frac{2018}{2}+...+\frac{2018}{2017}+\frac{2018}{2018}-2018\)

\(=2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)

\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}=\frac{1}{2018}\)

2S=2.(2^1+2^2+2^3+2^4+...+2^2017+2^2018)=2^2+...............2^2019

2S-S=2^2+...............2^2019-(2^1+2^2+2^3+2^4+...+2^2017+2^2018)

S=2^2019-2

\(S=2^1+2^2+2^3+2^4+..+2^{2017}+2^{2018}\)

\(S=2^{2018}+2^{2017}+....+2^4+2^3+2^2+2^1\)

\(2S=2^{2019}+2^{2018}+....+2^5+2^4+2^3+2^2\)

\(\Rightarrow2S-S=2^{2019}-2\)

hay \(S=2^{2019}-2\)

`Answer:`

\(T=\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)

\(\Leftrightarrow2T=2+\frac{3}{2}+\frac{4}{2^2}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)

\(\Leftrightarrow2T-T=2+\left(\frac{3}{2}-\frac{2}{2}\right)+\left(\frac{4}{2^2}-\frac{4}{2^2}\right)+...+\left(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

\(\Leftrightarrow2T-T=2+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

Ta đặt \(V=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(\Rightarrow T=2+V-\frac{2017}{2^{2016}}\text{(*)}\)

\(\Leftrightarrow2V=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)

\(\Leftrightarrow2V-V=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)

\(\Leftrightarrow2V-V=1-\frac{1}{2^{2015}}\text{(**)}\)

Từ (*)(**)\(\Rightarrow T=2+\left(1-\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)

\(\Leftrightarrow T=3-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)

`=>T<3`

\(M=\dfrac{3}{1\times2}+\dfrac{3}{2\times3}+\dfrac{3}{3\times4}+...+\dfrac{3}{2015\times2016}+\dfrac{3}{2016\times2017}\)
\(=3\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2015\times2016}+\dfrac{1}{2016\times2017}\right)\)
\(=3\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)
\(=3\times\left(1-\dfrac{1}{2017}\right)\)
\(=3\times\dfrac{2016}{2017}\)
\(=\dfrac{6048}{2017}\)
#DatNe

Đặt A = \(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2010.2012+2}{2011^2}+\frac{2015.2017+2}{2016^2}\)

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

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

\(=\frac{2^2+1}{2^2}+\frac{3^2+1}{3^2}+...+\frac{2016^2+1}{2016^2}\)

\(=\left(1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\right)\)(2015 hạng tử 1)

\(=2015+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2016.2016}\right)\)

 \(< 2015+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)

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

= 2015 + 1 + 1/2016

= 2016 + 1/2016 < 2017 

=> A < 2017 (ĐPCM)