Toán Lớp 6: Tính D = 1+1/3+1/3^2+1/3^3+…+1/3^100
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Toán Lớp 6: Tính D = 1+1/3+1/3^2+1/3^3+…+1/3^100
Comments ( 2 )
Lời giải và giải thích chi tiết:
D = 1+1/3+1/3^2+1/3^3+…+1/3^100
3D= (1+1/3+1/3^2+1/3^3+…+1/3^100).3
3D=3+1+1/3+1/{3^2}+…+1/{3^99}
3D-D=(3+1+1/3+1/{3^2}+…+1/{3^99})- (1+1/3+1/3^2+1/3^3+…+1/3^100)
2D=3+1+1/3+1/{3^2}+…+1/{3^99}- 1-1/3-1/3^2-1/3^3-…-1/3^100)
2D=3-1/{3^100}
D=\frac{3-\frac{1}{3^{100}}}{2}
Vậy D=\frac{3-\frac{1}{3^{100}}}{2}
#Math
Giải đáp:
D = ( 3^101 – 1 )/( 2 . 3^100 )
Lời giải và giải thích chi tiết:
D = 1 + 1/3 + 1/(3^2) + 1/(3^3) + … +1/(3^100)
=> 3D = 3 + 1 + 1/3 + 1/(3^2) + 1/(3^3) + … +1/(3^99)
=> 3D – D = ( 3 + 1 + 1/3 + … +1/(3^99) ) – ( 1 + 1/3 + 1/(3^2) +… +1/(3^100) )
=> 2D = 3 + 1 + 1/3 + … +1/(3^99) – 1 – 1/3 – 1/(3^2) – … – 1/(3^100)
=> 2D = 3 + ( 1 – 1 ) + ( 1/3 – 1/3 ) + … + ( 1/(3^99) – 1/(3^99) ) – 1/( 3^100 )
=> 2D = 3 – 1/( 3^100 )
=> 2D = ( 3^101 – 1 )/( 3^100 )
=> D = ( 3^101 – 1 )/( 3^100 ) . 1/2
=> D = ( 3^101 – 1 )/( 2 . 3^100 )
Vậy D = ( 3^101 – 1 )/( 2 . 3^100 )