Question

[NiNi love bebi Thảo My nhìu]

1:a) Tính 

\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

  b)  CMR : A<3

Answer 1

[Minh Triều]

a)\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

=\(\frac{9.4}{1.3.5}+\frac{9.4}{3.5.7}+\frac{9.4}{5.7.9}+...+\frac{9.4}{25.27.29}\)

=\(9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

=\(9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

=\(9.\left(\frac{1}{3}-\frac{1}{27.29}\right)=9.\left(\frac{1}{3}-\frac{1}{783}\right)=9.\left(\frac{261}{783}-\frac{1}{783}\right)=9.\frac{260}{783}\)

=\(\frac{260}{87}\)

b)

ta có: \(3=\frac{261}{87}>\frac{260}{87}\)

vậy A<3

Answer 2

[Phạm Quoc Trung]

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