Question

Cho A= (1+ 1/3) . (1+1/8) . (1+1/15)...(1+1/9999)

So sánh A với 2

Nhanh mk tk nha, cám ơn

Answer 1

ta có:

\(A=\left(1+\frac{1}{3}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right)....\left(1+\frac{1}{9999}\right)\)

\(A=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{10000}{9999}=\frac{2^2}{1.3}.\frac{3^2}{2.4}....\frac{100^2}{99.101}\)

\(A=\frac{\left(2.3.4.5....100\right)}{1.2.3.4....99}.\frac{\left(2.3.4...100\right)}{3.4.5..101}\)

\(A=\frac{100}{1}.\frac{2}{101}=\frac{200}{101}< \frac{202}{101}=2\)

\(\Rightarrow A< 2\)

Answer 2

nếu đúng k giúp mình nhé

Answer 3

Hi cám ơn bn nha

Answer 4

A = ( 1 + 1/3 ).( 1 + 1/8 ).( 1 + 1/15) ... ( 1 + 1/9999 )

A = 4/3 . 9/8 . 16/15 ..... 10000/9999

A = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\),\(\frac{4.4}{3.5}\)...\(\frac{100.100}{99.101}\)

A = \(\frac{2}{1}\).\(\frac{2}{3}\).\(\frac{3}{2}\).\(\frac{3}{4}\)....\(\frac{100}{99}\).\(\frac{100}{101}\)

A = ( \(\frac{2}{1}\).\(\frac{3}{2}\)....\(\frac{100}{99}\)).( \(\frac{2}{3}\).\(\frac{3}{4}\)...\(\frac{100}{99}\))

A = 100 . \(\frac{2}{99}\) = \(\frac{200}{99}\)