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

Toán Lớp 7: tính dãy sau. C=1^3+2^3+3^3+…+n^3

Home/ Toán Lớp 7: tính dãy sau. C=1^3+2^3+3^3+…+n^3

Toán Lớp 7: tính dãy sau. C=1^3+2^3+3^3+…+n^3

Tháng Tám 17, 2022 Toán học 1 Comment 0 views

Toán Lớp 7: tính dãy sau. C=1^3+2^3+3^3+…+n^3

Comments ( 1 )

  1. haiyemy
    haiyemy
    17 Tháng Tám, 2022 at 4:52 sáng
    Reply
    Giải đáp:
    $C=\dfrac{17.(n-1).n.(n+1).(n+2)}{12}$
    Lời giải và giải thích chi tiết:
    $C=1^3+2^3+3^3\ +\,.\!.\!.+\ n^3 \\\Rightarrow C=1.1.1+2.2.2+3.3.3\ +\,.\!.\!.+\ n.n.n \\\Rightarrow C=1.1.(2-1)+2.2.(3-1)+3.3.(4-1)\ +\,.\!.\!.+\ n.n.(n+1-1) \\\Rightarrow C=1.1.2-1.1+2.2.3-2.2+3.3.4-3.3+…+n.n.(n+1)-n.n \\\Rightarrow C=\bigg(\underbrace{1.1.2+2.2.3+3.3.4+…+n.n.(n+1)}_{\large C’}\bigg)-\bigg(\underbrace{1.1+2.2+3.3+…+n.n}_{\large C”}\bigg) \\\Rightarrow C’=1.1.2+2.2.3+3.3.4+…+n.n.(n+1) \\\Rightarrow C’=(0+1).1.2+(1+1).2.3+(2+1).3.4+…+(n-1+1).n.(n+1) \\\Rightarrow C’=0.1.2+1.2+1.2.3+2.3+2.3.4+3.4+…+(n-1).n.(n+1)+n.(n+1) \\\Rightarrow C’=\bigg(\underbrace{0.1.2+1.2.3+2.3.4+…+(n-1).n.(n+1)}_{\large C”’}\bigg)+\bigg(\underbrace{1.2+2.3+3.4+…+n.(n+1)}_{\large C””}\bigg) \\\Rightarrow C”’=0.1.2+1.2.3+2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1)\\\Rightarrow 4C”’=1.2.3.4+2.3.4.4\ +\,.\!.\!.+\ (n-1).n.(n+1).4\\\Rightarrow 4C”’=1.2.3.4+2.3.4.(5-1)\ +\,.\!.\!.+\ (n-1).n.(n+1).[(n+2)-(n-2)]\\\Rightarrow 4C”’=1.2.3.4+2.3.4.5-1.2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1).(n+2)-(n-2).(n-1).n.(n+1)\\\Rightarrow 4C”’=(1.2.3.4+2.3.4.5\ +\,.\!.\!.+\ (n-1).n.(n+1).(n+2))-(1.2.3.4+2.3.4.5\ +\,.\!.\!.+\ (n-2).(n-1).n.(n+1))\\\Rightarrow 4C”’=(n-1).n.(n+1).(n+2)\\\Rightarrow C”’=\dfrac{(n-1).n.(n+1).(n+2)}{4}\\\Rightarrow C””=1.2+2.3+3.4\ +\,.\!.\!.+\ n.(n+1)\\\Rightarrow 3C””=1.2.3+2.3.3+3.4.3\ +\,.\!.\!.+\ n.(n+1).3\\\Rightarrow 3C””=1.2.3+2.3.(4-1)+3.4.(5-2)\ +\,.\!.\!.+\ n. (n+1).[(n+2)-(n-1)]\\\Rightarrow 3C””=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4\ +\,.\!.\!.+\ n.(n+1).(n+2)-(n-1).n.(n+1)\\\Rightarrow 3C””=(1.2.3+2.3.4+3.4.5\ +\,.\!.\!.+\ n.(n+1).(n+2))-(1.2.3+2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1)\\\Rightarrow 3C””=n.(n+1).(n+2)\\\Rightarrow C””=\dfrac{n.(n+1).(n+2)}{3}\\\Rightarrow C’=\dfrac{(n-1).n.(n+1).(n+2)}{4}+\dfrac{n.(n+1).(n+2)}{3}\\\Rightarrow C’=\dfrac{7.(n-1).n.(n+1).(n+2)}{12}\\\Rightarrow C”=1.1+2.2+3.3\ +\,.\!.\!.+\ n.n\\\Rightarrow C”=1+2.(1+1)+3.(2+1)\ +\,.\!.\!.+\ n.(n-1+1)\\\Rightarrow C”=1+1.2+2+2.3+3\ +\,.\!.\!.+\ (n-1).n+n\\\Rightarrow C”=\bigg(\underbrace{1.2+2.3\ +\,.\!.\!.+\ (n-1).n}_{\large C,}\bigg)+\bigg(1+2+3\ +\,.\!.\!.+ \ n\bigg)\\\Rightarrow C,=1.2+2.3\ +\,.\!.\!.+\ (n-1).n\\\Rightarrow 3C,=1.2.3+2.3.3\ +\,.\!.\!.+\ (n-1).n.3\\\Rightarrow 3C,=1.2.3+2.3.(4-1)\ +\,.\!.\!.+\ (n-1).n.[(n+1)-(n-2)]\\\Rightarrow 3C,=1.2.3+2.3.4-1.2.3\ +\,.\!.\!.+\ (n-1).n.(n+1)-(n-2).(n-1).n\\\Rightarrow 3C,=(1.2.3+2.3.4\ +\,.\!.\!.+\ (n-1).n.(n+1))-(1.2.3+2.3.4\ +\,.\!.\!.+\ (n-2).(n-1).n\\\Rightarrow 3C,=(n-1).n.(n+1)\\\Rightarrow C,=\dfrac{(n-1).n.(n+1)}{3}\\\Rightarrow C”=\dfrac{(n-1).n.(n+1)}{3}+\dfrac{n.(n+1)}{2}\\\Rightarrow C”=\dfrac{5.(n-1).n.(n+1)}{6}\\\Rightarrow C=\dfrac{7.(n-1).n.(n+1).(n+2)}{12}+\dfrac{5.(n-1).n.(n+1)}{6}\\\Rightarrow C=\dfrac{17.(n-1).n.(n+1).(n+2)}{12}$

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

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