Question

[Karina]

  các bạn giải bài 9 cho mình nhé mình cần gấp

Answer 1

[Nguyễn Thị Thương Hoài]

Bài 9

a; A = (1 + \(\dfrac{1}{3}\))\(\times\)(1 + \(\dfrac{1}{8}\))\(\times\)(1 + \(\dfrac{1}{15}\))\(\times\)...\(\times\)(1 + \(\dfrac{1}{9999}\))

    A = \(\dfrac{3+1}{3}\)\(\times\)\(\dfrac{8+1}{8}\)\(\times\)\(\dfrac{15+1}{15}\)\(\times\)...\(\times\)\(\dfrac{9999+1}{9999}\)

   A  = \(\dfrac{4}{3}\)\(\times\)\(\dfrac{9}{8}\)\(\times\)\(\dfrac{16}{15}\)\(\times\)...\(\times\)\(\dfrac{10000}{9999}\)

A = \(\dfrac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)

A = \(\dfrac{2\times100}{1\times101}\)

A = \(\dfrac{200}{101}\)

b; B = (1 - \(\dfrac{1}{4}\))\(\times\) (1 - \(\dfrac{1}{9}\)) \(\times\) (1 - \(\dfrac{1}{16}\))\(\times\)...\(\times\)(1 - \(\dfrac{1}{10000}\))

    B = \(\dfrac{4-1}{4}\) \(\times\) \(\dfrac{9-1}{9}\) \(\times\) \(\dfrac{16-1}{16}\) \(\times\) \(\dfrac{10000-1}{10000}\)

    B = \(\dfrac{3}{4}\)\(\times\) \(\dfrac{8}{9}\)\(\times\)\(\dfrac{15}{16}\)\(\times\)...\(\times\) \(\dfrac{9999}{10000}\)

   B = \(\dfrac{1\times3}{2\times2}\)\(\times\)\(\dfrac{2\times4}{3\times3}\)\(\times\)\(\dfrac{3\times5}{4\times4}\)\(\times\)...\(\times\)\(\dfrac{99\times101}{100\times100}\)

  B = \(\dfrac{1\times101}{2\times100}\)

  B = \(\dfrac{101}{200}\)