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Câu 24:
\(D=(\frac{-1}{7})^0+(\frac{-1}{7})^1+(\frac{-1}{7})^2+...+(\frac{-1}{7})^{2017}\)
\(\frac{-1}{7}D=(\frac{-1}{7})^1+(\frac{-1}{7})^2+(\frac{-1}{7})^3+...+(\frac{-1}{7})^{2018}\)
\(\frac{-1}{7}D-D=(\frac{-1}{7})^{2018}-(\frac{-1}{7})^0\)
\(\frac{-8}{7}D=\frac{1}{7^{2018}}-1=\frac{1-7^{2018}}{7^{2018}}\)
\(D=\frac{7^{2018}-1}{8.7^{2017}}\)
Không đáp án nào đúng.
Câu 25:
\(E=\frac{-1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)
\(\frac{1}{3}E=\frac{-1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{51}}-\frac{1}{3^{52}}\)
\(\Rightarrow E+\frac{1}{3}E=\frac{-1}{3}+\frac{-1}{3^{52}}\)
$\frac{4}{3}E=-\frac{3^{51}+1}{3^{52}}$
$E=-\frac{3^{51}+1}{4.3^{51}}$
Đáp án D.
Câu 26:
\(F=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+....+\frac{100}{2^{100}}\)
\(2F=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
\(F=2F-F=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(F+\frac{100}{2^{100}}=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2(F+\frac{100}{2^{100}})=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2(F+\frac{100}{2^{100}})-(F+\frac{100}{2^{100}})=2-\frac{1}{2^{99}}\)
\(F=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}=\frac{2^{101}-102}{2^{100}}\)
Đáp án D.
Câu 27:
\(G=\frac{3}{5}+\frac{3}{5^4}+\frac{3}{5^7}+...+\frac{3}{5^{100}}\)
\(\frac{1}{5^3}G=\frac{3}{5^4}+\frac{3}{5^7}+...+\frac{3}{5^{103}}\)
\(G-\frac{1}{5^3}G=\frac{3}{5}-\frac{3}{5^{103}}=\frac{3.5^{102}-3}{5^{103}}\)
\(\frac{124}{5^3}G=\frac{3.5^{102}-3}{5^{103}}\Rightarrow G=\frac{3.5^{102}-3}{124.5^{100}}\)
Không có đáp án đúng.
Câu 28:
\(A=2^2+(2^2+2^3+2^4+...+2^{50})\)
\(2A=2^3+(2^3+2^4+2^5+...+2^{51})\)
\(2A-A=(2^3-2^2)+(2^{51}-2^2)=2^{51}\)
\(A=2^{51}\)
Đáp án B.
Câu 29:
$A=1+3+3^2+3^3+...+3^{100}$
$3A=3+3^2+3^3+...+3^{101}$
$3A-A=3^{101}-1$
$2A=3^{101}-1$
$2A+1=3^{101}=3^n$
$\Rightarrow n=101$
