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

Toán Lớp 7: cho Q= 4/1x2x3 + 7/2x3x4 + – + 151/50x51x52 a.tìm dạng tổng quát của Q b.tính Q

Home/ Toán Lớp 7: cho Q= 4/1x2x3 + 7/2x3x4 + – + 151/50x51x52 a.tìm dạng tổng quát của Q b.tính Q

Toán Lớp 7: cho Q= 4/1x2x3 + 7/2x3x4 + – + 151/50x51x52 a.tìm dạng tổng quát của Q b.tính Q

Tháng Mười 16, 2022 Toán học 1 Comment 0 views

Toán Lớp 7: cho Q= 4/1x2x3 + 7/2x3x4 + …. + 151/50x51x52
a.tìm dạng tổng quát của Q
b.tính Q

Comments ( 1 )

  1. maitrucloan
    maitrucloan
    16 Tháng Mười, 2022 at 12:43 chiều
    Reply
    Giải đáp: $Q=\dfrac{8975}{5304}$
    Lời giải và giải thích chi tiết:
    a.Dạng tổng quát số hạng của $Q$ là:
    $$\dfrac{3n+1}{n\cdot \left(n+1\right)\cdot \left(n+2\right)}$$
    b.Ta có:
    $\dfrac{3n+1}{n\cdot \left(n+1\right)\cdot \left(n+2\right)}$
    $=\dfrac{3n}{n\cdot \left(n+1\right)\cdot \left(n+2\right)}+\dfrac{1}{n\cdot \left(n+1\right)\cdot\left(n+2\right)}$
    $=3\cdot \dfrac{1}{\left(n+1\right)\cdot \left(n+2\right)}+\dfrac12\cdot \dfrac{2}{n\cdot \left(n+1\right)\cdot\left(n+2\right)}$
    $=3\cdot \dfrac{\left(n+2\right)-\left(n+1\right)}{\left(n+1\right)\cdot \left(n+2\right)}+\dfrac12\cdot\dfrac{\left(n+2\right)-n}{n\cdot \left(n+1\right)\cdot\left(n+2\right)}$
    $=3\left(\dfrac1{n+1}-\dfrac1{n+2}\right)+\dfrac12\left(\dfrac1{n\left(n+1\right)}-\dfrac1{\left(n+1\right)\left(n+2\right)}\right)$
    Ta có:
    $Q=\dfrac4{1\cdot 2\cdot 3}+\dfrac7{2\cdot 3\cdot 4}+…+\dfrac{151}{50\cdot 51\cdot 52}$
    $\to Q=\left(3\left(\dfrac1{1+1}-\dfrac1{1+2}\right)+\dfrac12\left(\dfrac1{1\left(1+1\right)}-\dfrac1{\left(1+1\right)\left(1+2\right)}\right)\right)+…+\left(3\left(\dfrac1{50+1}-\dfrac1{50+2}\right)+\dfrac12\left(\dfrac1{50\left(50+1\right)}-\dfrac1{\left(50+1\right)\left(50+2\right)}\right)\right)$
    $\to Q=\left(3\left(\dfrac1{2}-\dfrac1{3}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{2\cdot 3}\right)\right)+…+\left(3\left(\dfrac1{51}-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{50\cdot 51}-\dfrac1{51\cdot 52}\right)\right)$
    $\to Q=3\left(\dfrac1{2}-\dfrac1{3}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{2\cdot 3}\right)+…+3\left(\dfrac1{51}-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{50\cdot 51}-\dfrac1{51\cdot 52}\right)$
    $\to Q=3\left(\dfrac12-\dfrac13+…+\dfrac1{51}-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{2\cdot 3}+….+\dfrac1{50\cdot 51}-\dfrac1{51\cdot 52}\right)$
    $\to Q=3\cdot \left(\dfrac12-\dfrac1{52}\right)+\dfrac12\left(\dfrac1{1\cdot 2}-\dfrac1{51\cdot 52}\right)$
    $\to Q=\dfrac{8975}{5304}$

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

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