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Câu a:
A = 1+ 3 + 3^2+ ..+ 3^100
3A = 3 + 3^2+ 3^3 + ...+ 3^101
3A - A = 3 + 3^2+ 3^3 + ...+ 3^101 - (1+ 3 + 3^2+ ..+ 3^100)
2A = 3 + 3^2 + 3^3 + ...+ 3^100 - 1 - 3 - 3^2-..-3^100
2A = (3^101-1)+(3^2-3^2)+..+(3^100-3^100)
2A = 3^101 - 1 + 0 + 0+ ...+ 0
2A = 3^101 - 1
A = (3^101 - 1)/2
Câu b:
A = 1+ 4 + 4^2+ ..+ 4^100
4A = 4 + 4^2+ 4^3 + ...+ 4^101
4A - A = 4 + 4^2+ 4^3 + ...+ 4^101 - (1+ 4 + 4^2+ ..+ 4^100)
4A - A= =4 + 4^2 + 4^3 + ...+ 4^100 - 1 - 4 - 4^2-..-4^100
3A = (4^101-1)+(4^2-4^2)+..+(4^100-4^100)
3A = 4^101 - 1 + 0 + 0+ ...+ 0
3A = 4^101 - 1
A = (4^101 - 1)/3
Ta có: \(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\)
\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(\Rightarrow3D-D=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{101}{3^{101}}\right)\)
\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow6D-2D=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{100}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(\Rightarrow4D=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(\Rightarrow4D< 3-\frac{203}{3^{100}}< 3\Rightarrow D< \frac{3}{4}\left(ĐPCM\right)\)
Gọi B=(1+3+3^2+3^3+..+3^100)
=>3B = 3^1 + 3^2 + 3^3 + 3^4 +...+ 3^100 + 3^101
=>3B - B = ( 3^1 + 3^2 + 3^3 + 3^4 +...+ 3^100 + 3^101 ) - ( 1 + 3 + 3^2 + 3^3 +...+ 3^100 )
=> 2B = 3^101 - 3
=>B= 3^101 - 3
_______
2
=>S= 3^101 - 3^101 - 3
_______
2

(1013-13).(1003-23)..................(23-1003).(13-1013)
=(1013-13).(1003-23).........(513-513).........(23-1003).(13-1013)
=(1013-13).(1003-23)...........0............(23-1003).(13-1013)
=0
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