S=3/2^0+3/2^1+....+3/2^2018

S=3/2.(2/2^0+2/2^1+....+2^2018)

đặt B=2/2^0+2/2^1+....+2^2018

2B=2.(2/2^0+2/2^1+....+2^2018)

2B=1+2/2^0+...+2/2^2017

2B-B=(1+2/2^0+...+2/2^2017)-(2/2^0+2/2^1+....+2^2018)

B=1-2^2018

S=3/2.1-2^2018=3/2^2018

B=2^2018-1 nha mink làm lộn

Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}=\frac{10^{2014}+10}{10^{2015}+10}=\frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}=\frac{10^{2013}+1}{10^{2014}+1}=A\)

\(\Rightarrow\)\(B< A\) hay \(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~ 

áp dụng tính chất

nếu a/b>1thì a/b<(a+n)/(b+n)

=)))))))))))))))))

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+.......+\frac{19}{9^2.10^2}\)

\(A=\frac{3}{1.4}+\frac{5}{4.9}+.......+\frac{19}{81.100}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+.......+\frac{1}{81}-\frac{1}{100}\)

\(A=1-\frac{1}{100}\)

\(A=\frac{99}{100}< \frac{100}{100}=1\)

\(\Rightarrow A< 1\)

mik nghĩ câu trả lời của nghĩa đúng nhưng mà 2 bước cuối phải thay bằng vì 1-^100 < 1 nên A < 1

 Ta có : \(A=\frac{3}{1^2+2^2}+\frac{5}{2^2+3^2}+\frac{7}{3^2+4^2}+...+\frac{19}{9^2+10^2}\)

\(\Leftrightarrow A=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(\Leftrightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(\Leftrightarrow A=1-\frac{1}{100}< 1\)

Vậy A < 1

\(C=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+....+\frac{99.100-1}{100!}\)

\(\Rightarrow C=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(\Rightarrow C=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(\Rightarrow C=\left(2+\frac{3.4}{4!}+\frac{4.5}{5!}+....+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{10!}\right)\)

\(\Rightarrow C=\left(2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)

\(\Rightarrow C=2-\frac{1}{99!}-\frac{1}{100!}< 2\Rightarrow C< 2\)

\(b,C=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{19}{9^2.10^2}\)

\(\Rightarrow C=\frac{3}{\left(1.2\right)\left(1.2\right)}+\frac{5}{\left(2.3\right)\left(2.3\right)}+...+\frac{19}{\left(9.10\right)\left(9.10\right)}\)

\(\Rightarrow C=\frac{3}{1.2}.\frac{1}{1.2}+\frac{5}{2.3}.\frac{1}{2.3}+....+\frac{19}{9.10}.\frac{1}{9.10}\)

\(\Rightarrow C=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{9}+\frac{1}{10}\right)\left(\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{90}\)

\(\Rightarrow C=1-\frac{1}{90}< 1\Rightarrow C< 1\)

câu 1hinhf như sai đề

Tớ nghĩ là S= 3⁰ + 3² + 3⁴ +3⁶ +...+ 3²⁰⁰²

^{thì đúng hơn}

sory. đề bài 1 là \(S=3^0+3^2+3^4+.....+3^{2002}\)

1,

a,   \(S=3^0+3^1+...+3^{2002}\)

\(\Leftrightarrow3S=3^1+3^2+...+3^{2003}\)

\(\Leftrightarrow3S-S=\left(3^1+3^2+...+3^{2003}\right)-\left(3^0+3^1+...+3^{2002}\right)\)

\(\Leftrightarrow2S=3^{2003}-1\)

\(\Leftrightarrow S=\frac{3^{2003}-1}{2}\)

D=\(\frac{1}{1^2}\)-\(\frac{1}{2^2}\)+\(\frac{1}{2^2}\)-\(\frac{1}{3^2}\)+...+\(\frac{1}{9^2}\)-\(\frac{1}{10^2}\)

D=\(\frac{1}{1^2}\)-\(\frac{1}{10^2}\)

D=\(1\)-\(\frac{1}{100}\)

D=\(\frac{99}{100}\)

a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)

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

\(=1-\frac{1}{2018}\)

\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)

b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2\left(1-\frac{1}{2019}\right)\)

\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2.\frac{2018}{2019}\)

\(=\frac{4036}{2019}\)

Phần c tương tự nha

a) \(\frac{1}{1.2}\) +  \(\frac{1}{2.3}\) + .......+  \(\frac{1}{2017.2018}\)

= 1 -  \(\frac{1}{2}\) + \(\frac{1}{2}\) -  \(\frac{1}{3}\) + .......+  \(\frac{1}{2017}\) -   \(\frac{1}{2018}\)

= 1 -  \(\frac{1}{2018}\) =  \(\frac{2017}{2018}\)

câu a) mik sửa đề một tí ko biết có đúng ko

câu b , c tương tự nhưng cần lấy tử ra chung 

a)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)nhóm 2 ra ngoài rồi làm như câu a

c)nhóm 4 ra rồi làm như câu a

Ta có : \(A=2+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

\(\Rightarrow A=2+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=2+\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=2+\frac{99}{100}=\frac{299}{100}\)

Ta có : A=\(2+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Rightarrow A=2+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow A=2+\left(1-\frac{1}{100}\right)\)

\(\Rightarrow A=2+\frac{99}{100}\)

\(\Rightarrow A=\frac{299}{100}\)

Can you k for me,Natsu drangeel!