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19 tháng 11 2018

\(A=3^{101}+3^{102}+3^{103}+...+3^{200}\)

\(3A=3^{102}+3^{103}+3^{104}+...+3^{201}\)

\(3A-A=\left(3^{102}+3^{103}+3^{104}+3^{201}\right)-\left(3^{101}+3^{102}+3^{103}+...+3^{201}\right)\)

\(2A=3^{201}-3^{101}\)

\(2A=3^{100}\)

\(\Rightarrow A=3^{100}:2\)

19 tháng 11 2018

\(A=3^{101}+3^{102}+3^{103}+...+3^{200}\)

\(A=3^{101}+3^{102}+3^{103}+3^{104}+...+3^{197}+3^{198}+3^{199}+3^{200}\)

\(A=3^{100}\left(3+3^2+3^3+3^4\right)+...+3^{196}\left(3+3^2+3^3+3^4\right)\)

\(A=120\left(3^{100}+...+3^{196}\right)⋮120\)

2 tháng 8 2023

\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)

\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)

\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)

\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)

\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)

\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)

\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)

\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)

\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)

24 tháng 9 2015

 

Ta thấy tổng trên có 50 số hạng .

Ta có:

1/101>1/150

1/102>1/150

...

1/149>1/150

1/150=1/150

=>1/101+1/102+...+1/149+1/150>1/150+1/150+...+1/150

                                                 ---50 số hạng 1/150-------

=>1/101+1/102+...+1/149+1/150>1/150.50

=>1/101+1/102+...+1/149+1/150>50/150

=>1/101+1/102+...+1/149+1/150>1/3

 

10 tháng 4 2017

\(A=2^{100}+2^{101}+2^{102}+...+2^{107}\)

\(A=2^{100}\left(1+2\right)+2^{102}\left(1+2\right)+...+2^{106}\left(1+2\right)\)

\(A=2^{100}.3+2^{102}.3+...+2^{106}.3\)

\(A=3\left(2^{100}+2^{102}+...+2^{106}\right)⋮3\)

2 tháng 8 2017

Ta có : A = 3 + 32 + 33 + 3+ ...... + 3100

=> A = (3 + 32 + 33 + 34) + ...... + (397 + 398 + 399 + 3100)

=> A = (3 + 32 + 33 + 34) + ...... + 396(3 + 32 + 33 + 34)

=> A = 120 + ..... + 396.120

=> A = 120(1 + .... + 396) chia hết cho 120

2 tháng 8 2017

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

  =\(\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

  =\(\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)

  =\(\left(3+3^2\right)\left(1+3^2+3^4+...+3^{98}\right)\)

  =\(12\left(1+3^2+3^4+...+3^{98}\right)\)

\(12⋮12\)=>\(12\left(1+3^2+3^4+...+3^{98}\right)⋮12\)

=>\(A⋮12\)

Vậy \(A⋮12\)