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\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
S= 2-1-2-3-4-....-201-202
= 2-(1+2+3+4+....+202)
=2- (202:2)( 202+1)
= 2- 101.203
S1 = 1-2+3-4+....+2017-2018
= (-1)+(-1)+....+(-1)
= (-1) x 1009
= -1009
Lần sau viết cái đề rõ rõ ra nhs!!!
a) \(A=2+2^2+2^3+................+2^{100}\)
\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)
\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)
\(\Rightarrow A=2^{101}-2\)
b) \(B=1+3+3^2+..................+3^{2009}\)
\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)
\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)
\(\Rightarrow2B=3^{2010}-1\)
\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)
c) \(C=4+4^2+4^3+................+4^n\)
\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)
\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)
\(\Rightarrow3C=4^{n+1}-4\)
\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)
\(4^{2017}:\left(4^{2014}+3\cdot4^{2014}\right)\)
\(=4^{2017}:4^{2014}\left(1+3\right)\)
\(=4^3\cdot4\)
\(=4^4\)
\(=256\)
4^2017 : ( 4^2014 + 3 . 4^2014 )
=\(\frac{\text{4^2017}}{\text{( 4^2014 + 3 . 4^2014 )}}\)
=\(\frac{\text{4^2017}}{\text{4^2017.4}}\)
=\(\frac{1}{4}\)