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

Toán Lớp 7: Chứng minh rằng : ` 3/( 1^2 . 2^2 ) + 5/( 2^2 . 3^2 ) + 7/(3^2 . 4^2) + … + 19/( 9^2 . 10^2 ) < 1 `

Toán Lớp 7: Chứng minh rằng :
3/( 1^2 . 2^2 ) + 5/( 2^2 . 3^2 ) + 7/(3^2 . 4^2) + … + 19/( 9^2 . 10^2 ) < 1

Comments ( 2 )

  1. đặt A = $\frac{3}{1^{2}.2^{2}}$+$\frac{5 }{ 2^{2}.3^{2}}$+$\frac{7}{3^{2}.4^{2}}$+…+$\frac{19}{9^{2}.10^{2}}$
    =$\frac{2^{2}-1^{2}}{1^{2}.2^{2}}$+$\frac{3^{2}- 2^{2} }{ 2^{2}.3^{2}}$+$\frac{4^{2}- 3^{2}}{3^{2}.4^{2}}$+…+$\frac{10^{2}- 9^{2} }{9^{2}.10^{2}}$
    =$\frac{1}{1^{2}}$ – $\frac{1}{2^{2}}$ + $\frac{1}{2^{2}}$ – $\frac{1}{3^{2}}$ + $\frac{1}{3^{2}}$ – $\frac{1}{4^{2}}$ + …+ $\frac{1}{9^{2}}$ – $\frac{1}{10^{2}}$ 
    =$\frac{1}{1^{2}}$ – $\frac{1}{10^{2}}$
    =$\frac{10^{2} – 1^{2}}{1^{2}. 10^{2}}$  
    =$\frac{100-1}{1.100}$ 
    =$\frac{99}{100}$ 
    DO 99<100
    $\frac{99}{100}$ < 1
    ⇒A < 1
    XIN 5*

  2. 3/(1^2*2^2) + 5/(2^2*3^2) + 7/(3^2*4^2) + … + 19/(9^2*10^2)
    = 1/(1^2) – 1/(2^2) + 1/(2^2) – 1/(3^2) + 1/(3^2) + …. + 1/(9^2) – 1/(10^2)
    = 1/(1^2) – 1/(10^2)
    =1 – 1/100 = 99/100 < 100/100 = 1

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