câu hỏi của bạn là gì

what is the 8th number in the sequence with pattern below

thì tăng lên 410 đơn vị nha

BẠN GHI ĐỀ ĐI RỒI CÓ GÌ MÌNH GÓP Ý CHO

Giá gốc của đôi giày:
\(50:20\times100=250\)(đô la)
Đáp số: 250 đô la

\(\dfrac{6}{16}-\dfrac{10}{64}=\dfrac{6\cdot4-10}{64}=\dfrac{14}{64}=\dfrac{7}{32}\)

\(C=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(3C=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(C+3C=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(4C=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(12C=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(4C+12C=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)

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

\(\Rightarrow C< \dfrac{3}{16}\)

gấp với mn

Có hình ko bn ơi?

\(C=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2C=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow2C+C=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\right)\)
\(\Rightarrow3C=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+2^{98}-2^{97}-....+2^2-2\)
            \(=2^{101}-\left(2^{100}-2^{100}\right)+\left(2^{99}-2^{99}\right)-\left(2^{98}-2^{98}\right)+...+\left(2^3-2^3\right)-\left(2^2-2^2\right)-2\)
            \(=2^{101}-2\)
\(\Rightarrow C=\dfrac{2^{101}-2}{3}\)