Question

CHỨNG MINH

a, 1.3.5.7.....197.199<\(\frac{101.102.102.....200}{1+2+2^2+2^3+...+2^{99}}\)

b, \(\frac{-1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2499}{2500}>\frac{-1}{\left(-7\right)^2}\)

Answer 1

a) dat A=1+2+2²+2³+...+2⁹⁹

2.A=2+2²+2³+2⁴+...+2¹⁰⁰

2.A-A= 2+2³+2⁴+...+2¹⁰⁰-(1+2+2²+2³+...+2⁹⁹)

A=2¹⁰⁰-1

----> 1.3.5.7...197.199<\(\frac{101.102.103....200}{2^{100}-1}\)

Dat B =1.3.5.7...197.199 

B=\(\frac{1.3.5.7....197.199...2.4.6.8....200}{2.4.6.8....200}\)

B= \(\frac{1.2.3.4.5....199.200}{2.4.6.8....200}\)

B=\(\frac{1.2.3.4.5......199.200}{2^{100}.\left(1.2.3.4...100\right)}\) ( tu 2 den 200 co 100 so hang nen duoc 2¹⁰⁰)

B =\(\frac{101.102.103....200}{2^{100}}\)

---->\(\frac{101.102.103....200}{2^{100}}

Answer 2

b> A= \(\frac{1.3.5.7....2499}{2.4.6.8....2500}\)  chon B=\(\frac{2.4.6.8...2500}{3.5.7.9...2501}\)

A.B = \(\frac{1.3.5.7....2499.2.4.6.8...2500}{2.4.6.8...2500.3.5.7.....2499.2501}=\frac{1}{2501}\)

Nhan xet 

\(\frac{1}{2}+\frac{1}{2}=1\)

\(\frac{2}{3}+\frac{1}{3}=1\)

vi 1/2 >1/3----> 1/2 <2/3

cm tuong tu ta se co A<B

---> A.A<A.B

---->A²<A.B

===> A² <\(\frac{1}{2501}