Toán 7 CMR $\dfrac{1}{2^2}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{4n-2}}-\dfrac{1}{2^{4n}}<\dfrac 12$

[Cute Boy]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!

ĐĂNG BÀI NGAY để cùng trao đổi với các thành viên siêu nhiệt tình & dễ thương trên diễn đàn.

giúp mik với CMR
[tex]1,B=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}<\frac{1}{2} 2,S=\frac{1}{2^{2}}-\frac{1}{2^{4}}+\frac{1}{2^{6}}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}<\frac{1}{2} 3,P=\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+...+\frac{1}{200^{2}}<\frac{1}{2}[/tex]

 

[Hạt Đậu nhỏ]

giúp mik với CMR
[tex]1,B=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}<\frac{1}{2} 2,S=\frac{1}{2^{2}}-\frac{1}{2^{4}}+\frac{1}{2^{6}}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}<\frac{1}{2} 3,P=\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+...+\frac{1}{200^{2}}<\frac{1}{2}[/tex]

2 câu còn lại làm tương tự

 

[besttoanvatlyzxz]

giúp mik với CMR
[tex]1,B=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}<\frac{1}{2} 2,S=\frac{1}{2^{2}}-\frac{1}{2^{4}}+\frac{1}{2^{6}}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}<\frac{1}{2} 3,P=\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+...+\frac{1}{200^{2}}<\frac{1}{2}[/tex]

1, [tex]B=\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{2005}}\\\\ => 3B=1+\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{2004}}\\\\ => 3B-B=1-\frac{1}{3^{2004}}\\\\ => B=\frac{1}{2}-\frac{1}{2.3^{2004}}<\frac{1}{2}\\\\\\ b, S=\frac{1}{2^{2}}-\frac{1}{2^{4}}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}\\\\ => 4S=1-\frac{1}{2^{2}}+\frac{1}{2^{4}}-...+\frac{1}{2^{4n}}-\frac{1}{2^{4n+2}}\\\\ => 4S+S=1-\frac{1}{2^{4n+2}}\\\\ => S=\frac{1}{5}-\frac{1}{5.2^{4n+2}} <\frac{1}{5}<\frac{1}{2}\\\\\\[/tex]
[tex]c,P=\frac{1}{2^{2}}+\frac{1}{4^{2}}+...+\frac{1}{200^{2}}\\\\ =\frac{1}{1.2^{2}}+\frac{1}{2^{2}.2^{2}}+...+\frac{1}{2^{2}.100^{2}}\\\\ => 4P=1+\frac{1}{2^{2}}+...+\frac{1}{100^{2}}\\\\ => 4P-P=1-(\frac{1}{102^{2}}+...+\frac{1}{200^{2}})\\\\ => P=\frac{1}{3}-3.(\frac{1}{102^{2}}+...+\frac{1}{200^{2}})<\frac{1}{3}<\frac{1}{2}[/tex]

 

[Cute Boy]

1, [tex]B=\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{2005}}\\\\ => 3B=1+\frac{1}{3}+\frac{1}{3^{2}}+...+\frac{1}{3^{2004}}\\\\ => 3B-B=1-\frac{1}{3^{2004}}\\\\ => B=\frac{1}{2}-\frac{1}{2.3^{2004}}<\frac{1}{2}\\\\\\ b, S=\frac{1}{2^{2}}-\frac{1}{2^{4}}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}\\\\ => 4S=1-\frac{1}{2^{2}}+\frac{1}{2^{4}}-...+\frac{1}{2^{4n}}-\frac{1}{2^{4n+2}}\\\\ => 4S+S=1-\frac{1}{2^{4n+2}}\\\\ => S=\frac{1}{5}-\frac{1}{5.2^{4n+2}} <\frac{1}{5}<\frac{1}{2}\\\\\\[/tex]
[tex]c,P=\frac{1}{2^{2}}+\frac{1}{4^{2}}+...+\frac{1}{200^{2}}\\\\ =\frac{1}{1.2^{2}}+\frac{1}{2^{2}.2^{2}}+...+\frac{1}{2^{2}.100^{2}}\\\\ => 4P=1+\frac{1}{2^{2}}+...+\frac{1}{100^{2}}\\\\ => 4P-P=1-(\frac{1}{102^{2}}+...+\frac{1}{200^{2}})\\\\ => P=\frac{1}{3}-3.(\frac{1}{102^{2}}+...+\frac{1}{200^{2}})<\frac{1}{3}<\frac{1}{2}[/tex]

3B - B = [tex]1-\frac{1}{3^{2005}}[/tex] mới đúng

 

[Vũ Lan Anh]

3B - B = [tex]1-\frac{1}{3^{2005}}[/tex] mới đúng

thay số đó vào vẫn đúng mà bạn!!
bạn ý nhầm có 1 xíu!!