Trang chủ » Hỏi đáp » Môn Toán b,cho M=1+3+3^2+3^3+…..+3^99+3^100 08/03/2024 b,cho M=1+3+3^2+3^3+…..+3^99+3^100
M=1+3+3^2+…+3^99+3^100 3M=3+3^2+3^3+…+3^100+3^101 3M-M=(3+3^2+3^3+…+3^100+3^101)-(1+3+3^2+…+3^99+3^100) 2M=3^101-3 M={3^101-3}/2 Trả lời
Ta có: M = 1 + 3 + 3^2 + 3^3 + … +3^(99) +3 ^(100) => 3M = 3 + 3^2 + 3^3 + 3^4 + …. + 3^(100) + 3^(101) => 3M – M = (3 + 3^2 + 3^3 + 3^4 + …. + 3^(100) + 3^(101)) – (1 + 3 + 3^2 + 3^3 + … +3^(99) +3 ^(100)) => 2M = 3^(101) – 1 => M = (3^(101) – 1)/2 Vậy M = (3^(101) – 1)/2 $#duong612009$ Trả lời
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