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    A  = 2 +  2² + 2³ +...+2⁹⁹ + 2¹⁰⁰

   2A  =       2² + 2³ +...+ 2⁹⁹ + 2¹⁰⁰ + 2¹⁰¹

2A - A =      2¹⁰¹ - 2

   A      = 2¹⁰¹ - 2

  B        =  1 + 5 + 5² + 5³ +...+ 5⁹⁹ + 5¹⁰⁰

5B        =       5 + 5² + 5³ +...+ 5⁹⁹ + 5¹⁰⁰ + 5¹⁰¹

5B - B   =  5¹⁰¹ - 1

   4B     = 5¹⁰¹ - 1

     B      = \(\dfrac{5^{101}-1}{4}\)

     B     = 

gọi biểu thức trên là A , ta có :

\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

từ đó ta được :

\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)

help mik với 

lx

ko có câu hỏi bn ơi

Lỗi

A=-1+(-1)+...+(-1) {có 50 số hạng}

=-1.50=-50

A=1-2+3-4+...+99-100       SSH=(100-1):1+1=100 Sh

=>A=(1-2)+(3-4)+....+(99-100)

vì chia thành cặp suy ra 100:2 =50 cặp

A=(-1)+(-1)+...(-1)

A=(-1).50

A=-50

bạn giúp mình trả lời chi tiết câu c nha cám ơn các ban

a tk

mik gủi 1 ý 1 lần nha

c.tk

thay c = a thôi