[TEX]A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{100}{2^{100}}[/TEX]
[TEX] \Rightarrow \frac{1}{2}A= \frac{1}{2}+ \frac{3}{2^4}+ \frac{4}{2^5}+...+ \frac{99}{2^{100}}+ \frac{100}{2^{101}}[/TEX]
[TEX]\Rightarrow A- \frac{1}{2} A= \left( 1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+ \frac{100}{2^{100}} \right) - \left( \frac{1}{2}+ \frac{3}{2^4}+ \frac{4}{2^5}+...+ \frac{99}{2^{100}}+ \frac{100}{2^{101}} \right) = \left( 1- \frac{1}{2} \right) + \frac{3}{2^3}+ \left( \frac{4}{2^4}- \frac{3}{2^4} \right) + \left( \frac{5}{2^5}- \frac{4}{2^5} \right) +...+ \left( \frac{100}{2^{100}}- \frac{99}{2^{100}} \right) - \frac{100}{2^{101}}[/TEX]
[TEX]= \frac{1}{2} + \frac{3}{2^3}+ \frac{1}{2^4}+ \frac{1}{2^5}+...+ \frac{1}{2^{100}}- \frac{100}{2^{101}}[/TEX].
Như vậy [TEX]\frac{1}{2} A = \frac{1}{2} + \frac{3}{2^3}+ \frac{1}{2^4}+ \frac{1}{2^5}+...+ \frac{1}{2^{100}}- \frac{100}{2^{101}}= \frac{1}{2}+ \frac{1}{2^2}+ \frac{1}{2^3}+ \frac{1}{2^4}+...+ \frac{1}{2^{100}}- \frac{100}{2^{101}}[/TEX].
Thực hiện tương tự như trên ta lại tìm được [TEX]\frac{1}{2}A= 1+ \frac{99}{2^{100}}+ \frac{100}{2^{101}} \Rightarrow A= 2+ \frac{99}{2^{99}}+ \frac{100}{2^{100}}[/TEX].
Mình xin làm tiếp từ đoạn[TEX]\frac{1}{2}+ \frac{1}{2^2}+ \frac{1}{2^3}+ \frac{1}{2^4}+...+ \frac{1}{2^{100}}- \frac{100}{2^{101}} [/TEX]
Ta chứng minh biểu thức
[TEX] \frac{1}{2}+ \frac{1}{2^2}+ \frac{1}{2^3}+\frac{1}{2^4}+ \frac{1}{2^5}+...+ \frac{1}{2^{100}}- \frac{100}{2^{n}}=1-\frac{1}{2^n}[/TEX]
=[TEX]1-(\frac{1}{2} - \frac{1}{2^2})+\frac{1}{2^3}+ \frac{1}{2^4}+ \frac{1}{2^5}+...+ \frac{1}{2^{100}}[/TEX]
=[TEX]1-(\frac{1}{2^2} - \frac{1}{2^3})+ \frac{1}{2^4}+ \frac{1}{2^5}+...+ \frac{1}{2^{100}}[/TEX]
=[TEX]1-(\frac{1}{2^3} - \frac{1}{2^4})+ \frac{1}{2^5}+ \frac{1}{2^6}+...+ \frac{1}{2^{100}}[/TEX]
........... Cứ như ta chứng minh được biểu thức trên.
Áp dụng ta có:
[TEX]\frac{1}{2}A=1-\frac{1}{2^{100}}-\frac{100}{2^{101}}\Rightarrow A=2-\frac{1}{2^{99}}-\frac{100}{2^101}[/TEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
