SSH:(20152-12):10+1=2015

(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152

-10+(-10)+(-10)+...+(-10)+20152

-10x(2015-1):2+20152=12

=> C=12

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

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

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

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

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

Ta có 1/2.2<1/1.2

         1/3.3<1/2.3

         1/4.4<1/3.4

  .........................

         1/20.20<1/19.20

=>1/2.2+1/3.3+1/4.4+...+1/20.20<1/1.2+1/2.3+1/3.4+...+1/19.20

=>A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20

=>A<1/1-1/20

=>A<20/20-1/20

=>A<19/20<20/20=1

=>A<1

 Vậy A<1

Giải:

A=1/2²+1/3²+1/4²+...+1/9²

Ta có:

1/2²<1/1.2

1/3²<1/2.3

1/4²<1/3.4

...

1/9²<1/8.9

⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9

A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9

A<1/1-1/9

A<8/9

Ta có:

1/2²>1/2.3

1/3²>1/3.4

1/4²>1/4.5

...

1/9²>1/9.10

⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10

A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10

A>1/2-1/10

A>2/5

Vậy 2/5<A<8/9 (đpcm)

Chúc bạn học tốt!

bạn ơi qua giúp mk với

mk viết nhầm 

A = 1 / 2² + 1 / 3² + 1 / 4² + ... + 1 / 80² mới đúng nhé

cái cuối là 1/80^2 ak