Câu hỏi của Lê Hoàng Minh

Question

Cho $A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}$. Chứng minh A < 2.

Thu Thao · Accepted Answer

$2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}$

=> $2A-A=A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}$

Đặt $B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}$

=> $2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}$

=> $B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}$

=> $A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}$

=> $A=2-\dfrac{102}{2^{100}}< 2$

Câu trả lời:

$2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}$

=> $2A-A=A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}$

Đặt $B=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}$

=> $2B=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{98}}$

=> $B=\dfrac{1}{2^2}-\dfrac{1}{2^{99}}$

=> $A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{100}{2^{100}}-\dfrac{1}{2^{99}}$

=> $A=2-\dfrac{102}{2^{100}}< 2$