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a
\(A=1+3+3^2+3^3+....+3^{100}\)
\(3A=3+3^2+3^3+3^4+.....+3^{101}\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(B=1-\frac{1}{2^{99}}\)
c
\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)
\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)
\(6C=5^{101}+1\)
\(C=\frac{5^{101}+1}{6}\)
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)
\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)
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c: \(=\dfrac{3}{2}-1-21=\dfrac{3}{2}-21=\dfrac{-39}{2}\)
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Bài 1:
a: \(2P=2^{101}-2^{100}+2^{98}-2^{97}+...+2^3-2^2\)
=>\(3P=2^{101}-2\)
hay \(P=\dfrac{2^{101}-2}{3}\)
b: \(5Q=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2+5\)
=>\(6Q=5^{101}+1\)
hay \(Q=\dfrac{5^{101}+1}{6}\)
\(=2^5+1-\dfrac{1}{8}=33-\dfrac{1}{8}=\dfrac{263}{8}\)