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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 6: BT1 ) Chứng minh rằng với mọi n ∈ N ta luôn có : 1 / 1 × 6 + 1/6 ×11+ 1/ 11 × 16 + – + 1 / ( 5n+ 1 ) × ( 5n+6) = n+1+5n+6

Toán Lớp 6: BT1 ) Chứng minh rằng với mọi n ∈ N ta luôn có :
1 / 1 × 6 + 1/6 ×11+ 1/ 11 × 16 + …. + 1 / ( 5n+ 1 ) × ( 5n+6) = n+1+5n+6
BT2 ) Chứng minh rằng :
A = 1/ 1.2.3 + 1/ 2.3.4 + 1/3.4.5 +…+ 1/18. 19.20 < 1/4

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    1/1.6+1/6.11+1/11.16+…+1/((5n+1).(5n+6))=(n+1)/(5n+6
    =>5/1.6+5/6.11+5/11.16+…+5/((5n+1).(5n+6))=(5n+5)/(5n+6)
    =>1-1/6+1/6-1/11+1/11-1/16+…+1/(5n+1)-1/(5n+6)=(5n+5)/(5n+6)
    =>1-1/(5n+6)=(5n+5)/(5n+6)
    =>(5n+6-1)/(5n+6)=(5n+5)/(5n+6)
    =>(5n+5)/(5n+6)=(5n+5)/(5n+6)
    Vậy 1/1.6+1/6.11+1/11.16+…+1/((5n+1).(5n+6))=(n+1)/(5n+6.
    2.
    A=1/(1.2.3)+1/(2.3.4)+1/(3.4.5)+…+1/(18.19.20)
    =>2A=2/(1.2.3)+2/(2.3.4)+2/(3.4.5)+…+2/(18.19.20)
    =>2A=(3-1)/(1.2.3)+(4-2)/(2.3.4)+(5-3)/(3.4.5)+…+(20-18)/(18.19.20
    =>2A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+…+1/18.19-1/19.20
    =>2A=1/1.2-1/19.20
    =>2A=1/2-1/19.20
    =>A=1/4-1/(2.19.20)<1/4
    =>A<1/4
    Vậy A<1/4.

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222-9+11+12:2*14+14 = ? ( )