Cho A=1+3+3mu2+3mu3+…+3mu100

2 bình luận về “Cho A=1+3+3mu2+3mu3+…+3mu100”

1. A=1+3+3mu2+3mu3+…+3mu100

   3A = 3 + 3² + 3³ + … + 3^101

   3A – A = ( 3 + 3² + 3³ + … + 3^101 ) – ( 1+3+3mu2+3mu3+…+3mu100 )

   2A = 3^101 – 1

   A = ( 3^101 – 1 ) : 2

   A = 3^101 – 1 /2

   @thaotrang2910

   Hoidap247

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2. Đặt :A=3+3^2+3^3+3^4+…3^99+3^100

   =>3A=3.(3^2+3^3+3^4…+3^99+3^100)

   =>3A=3^2+3^3+3^4+…+3^99+3^100+3^101

   =>3A-A=(3^2+3^3+3^4…+3^99+3^100+3^101)-(3+3^2+3^3+3^4…+3^99+3^100)

   =>2A=1+3^101-3

   =>A=(3^101-3)/2

   #vat

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