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6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
a)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}\)
P/s: Cj chỉ biết làm ý a thôi nhé! Có j ko hiểu cmt nhé!
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow\dfrac{1}{2}\left(1-\dfrac{1}{x+2}\right)=\dfrac{8}{17}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{8}{17}:\dfrac{1}{2}=\dfrac{16}{17}\)
\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{16}{17}=\dfrac{1}{17}\)
\(\Rightarrow x+2=17\rightarrow x=15\)
Vậy x = 15
\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
a, \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow98\left(2x+1\right)=99.2x\)
\(\Leftrightarrow2x=98\Rightarrow x=49\)
b: Đặt \(A=1-3+3^2-3^3+...+\left(-3\right)^x\)
\(=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^x\)
\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)
\(\Leftrightarrow-3A-A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}-...-1\)
\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)
\(\Leftrightarrow A=\dfrac{\left(-3\right)^{x+1}-1}{-4}=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)
\(\Leftrightarrow\dfrac{-\left(-3\right)^{x+1}+1}{4}=\dfrac{3^{2012}-1}{2}\)
\(\Leftrightarrow-\left(-3\right)^{x+1}+1=2\cdot3^{2012}-2\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=2\cdot3^{2012}-3\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=3\left(2\cdot3^{2011}-1\right)\)
\(\Leftrightarrow-\left(-3\right)^x=2\cdot3^{2011}-1\)
=>x=2010
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(A=1-\dfrac{1}{99}\)
\(A=\dfrac{98}{99}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\)
\(B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(B=\dfrac{22}{45}\)
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+.........+\dfrac{2}{97.99}\)
\(\Leftrightarrow A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow A=1-\dfrac{1}{99}=\dfrac{98}{99}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.......+\dfrac{2}{8.9.10}\)
\(\Leftrightarrow B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+......+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(\Leftrightarrow B=\dfrac{1}{2}-\dfrac{1}{90}\)
\(\Leftrightarrow B=\dfrac{22}{45}\)