Question

\(A=\frac{^{3^2}}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+...+\frac{3^2}{97\cdot100}\)  Tính Nhanh

 

Answer 1

1/3.A=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\)

=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{97}-\frac{1}{100}\)

=\(1-\frac{1}{100}\)

=\(\frac{99}{100}\)

=>A=\(\frac{99}{100}:\frac{1}{3}\)

=\(\frac{297}{100}\)

Answer 2

\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)

\(A=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=3.\left(1-\frac{1}{100}\right)\)

\(A=3.\frac{99}{100}=\frac{297}{100}\)

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