a, \(5S=5^2+5^3+...+5^{2017}\)

\(5S-S=5^{2017}-5\)

\(S=\frac{5^{2017}-5}{4}\)

b,\(3S=3^2+3^3+...+3^{101}\)

\(3S-S=3^{101}-3\)

\(S=\frac{3^{101}-3}{2}\)

c, \(3S=3-3^2+3^3-...-3^{2016}\)

\(3S+S=1-3^{2016}\)

\(4S=1-3^{2016}\)

\(S=\frac{1-3^{2016}}{4}\)

b, 3S = 3^2+3^3+.....+3^101

2S=3S-S=(3^3+3^3+.....+3^101)-(3+3^2+....+3^100) = 3^101-3

=> S = (3^101-3)/2

Tk mk nha

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

\(< \frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{1}+\frac{1}{1}=2\)

\(\Rightarrow\)\(A< 2\left(đpcm\right)\)

chúc bạn học tốt!!!

Bài 6 :

 2S = 6 + 3 + 3/2 + ... + 3/2^8

 2S = 6 - 3/2^9 + S

   S = 6 - 3/2^9

  Vậy S = 6 - 3/2^9

Bài 7 :

  Ta có : 

    A = 1/1 + 1/2^2 + 1/3^2 + ... + 1/50^2 < 1 + 1/(1x2) + 1/(2x3) + ... + 1/(49x50) = 1 + 1 - 1/50 < 1 + 1 = 2

  =)  A < 2

   Vậy A < 2

Bài 8 :

  Do A = 1 + 2/(2015^2014 - 1 ) và B = 1 + 2/(2015^2014 - 3 ) mà 2/(2015^2014 -1) < 2/(2015^2014 - 3 )

 =) A < B

   Vậy A < B

Bài 9:

  Do 196/197 > 196/(197+198) và 197/198 > 197/(197+198)

  =)  A > B

   Vậy A > B

A = 2015 + 2015² + ... + 2015²⁰

2014.A = 2015.A - A = (2015² + 2015³ + ... + 2015²¹) - (2015 + 2015² + ... + 2015²⁰) = 2015²¹ - 2015 

=> A = \(\frac{2015^{21}-2015}{2014}\)

B = 3 + 3² + ... + 3⁹⁹

2.B = 3.B - B = (3² + 3³ + ... + 3¹⁰⁰) - (3 + 3² + ... + 3⁹⁹) = 3¹⁰⁰ - 3

=> B = \(\frac{3^{100}-3}{2}\)

\(A=2-2^2+2^3-2^4+...+2^{2015}\)

\(2A=2^2-2^3+2^4-2^5+...+2^{2016}\)

\(2A+A=\left(2^2-2^3+2^4-2^5+...+2^{2016}\right)-\left(2-2^2+2^3-2^4+...+2^{2015}\right)\)

\(3A=2^2-2^3+2^4-2^5+...+2^{2016}+2-2^2+2^3-2^4+...+2^{2015}\)

\(3A=2^{2016}+2\)

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

Vậy \(A=\frac{2^{2016}+2}{3}\)

mở câu hỏi tương tự nhìn rồi dua vào

Bài 1 : 

Ta có :

\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)

Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)

Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)

Vậy \(A>B\)

Bài 2 :

Ta có :

\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)

\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)

\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)

\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)

Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên  \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)

Nên : \(M>4\)

Vậy \(M>4\)

Bài 3 : 

Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)

Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)

\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)

\(\Rightarrow A< \frac{3}{4}\)

Vậy \(A< \frac{3}{4}\)

Bài 4 :

\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow A=\frac{1008}{2017}\)

Vậy \(A=\frac{1008}{2017}\)

\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)

\(1-\frac{1}{x+2}=\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)

\(\Rightarrow x+2=2017\)

\(\Rightarrow x=2017-2=2015\)

Vậy \(x=2015\)

a) \(S=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2014\right)+2015\)

\(\Leftrightarrow S=\left(1-2\right)+\left(3-4\right)+....+\left(2013-2014\right)+2015\)

Vì từ 1 đến 2014 có 2014 số hạng => có 1007 cặp => Có 1007 cặp -1 và số 2015

\(\Rightarrow S=\left(-1\right)\cdot1007+2015\)

<=>S=-1007+2015

<=> S=1008