Cho B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh B<1
Những câu hỏi liên quan
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
chứng minh B <1
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)
\(B=1-\frac{1}{2^{99}}< 1\left(đpcm\right)\)
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B= \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh B<1
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh: B<1
B = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
B = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
2B = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
2B - B = \(1-\frac{1}{2^{99}}\)
=> B = \(1-\frac{1}{2^{99}}
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Tính \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+......+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\) ta được B=
Cho B=\(1+\frac{-1}{2}+\left(\frac{-1}{2}\right)^2+\left(\frac{-1}{2}\right)^3\) + \(\left(\frac{-1}{2}\right)^4+\left(\frac{-1}{2}\right)^5+........+\left(\frac{-1}{2}\right)^{98}+\left(\frac{-1}{2}\right)^{99}\)
Chứng minh B < \(\frac{2}{3}\)
MÌNH ĐANG CẦN GẤP NÊN BẠN NÀO ĐÓ GHI BÀI GIẢI RA HỘ MÌNH LUÔN NHÉ, CẢM ƠN BẠN TRƯỚC
Tính \(T=\left(\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right)X\left(\frac{1}{99}+\frac{2}{98}+...+\frac{98}{2}\right)-\left(\frac{1}{99}+\frac{2}{98}+..+\frac{99}{1}\right)X\left(\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}\right)\)
Tính B = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
Còn thiếu mũ 99 ở cuối cùng nha
Tính \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
bn chắc đề đúng chứ?chổ (1/2)^99 đó,2 cái liền hả?
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\(\frac{1}{2}\)+\(\frac{1^2}{2^2}\)+\(\frac{1^3}{2^3}\)+...+\(\frac{1^{98}}{2^{98}}\)+\(\frac{1^{99}}{2^{99}}\)
=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{99}}\)
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{98}}\)-\(\frac{1}{2^{99}}\) Còn lại tự làm nhá kết quả cuối cùng là 299-1/299
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Xem thêm câu trả lời
Cho \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
CMR : B < 1