数列分组求和Sn=(3+2)+(3^2 + 3×2 + 2^2)+…+〔3^n + 3^(n-1)×2 + 3^(n-2

1个回答

  • Sn=(3^2-2^2)/(3-2)+(3^3-2^3)/(3-2)+…+[3^(n+1)-2^(n+1)]/(3-2)…①

    =(3^2-2^2)+(3^3-2^3)+…+[3^(n+1)-2^(n+1)]

    =[3^2+3^3+…+3^(n+1)]-[2^2+2^3+…+2^(n+1)]

    =(3^n-1)*3^2/(3-1)-(2^n-1)*2^2/(2-1)…②

    =1/2*[3^(n+2)-3^2]-[2^(n+2)-2^2]

    =1/2*[3^(n+2)-9]-1/2*2*[2^(n+2)-4]

    =1/2*{3^(n+2)-9-2*[2^(n+2)-4]}

    =1/2*[3^(n+2)-2^(n+3)-1]

    附公式

    ①x^n-y^n=(x-y)[x^(n-1)+x^(n-2)*y+x^(n-3)*y^2+…+x*y^(n-2)+y^(n-1)](n为正整数)

    变形得此步所用的x^(n-1)+x^(n-2)*y+x^(n-3)*y^2+…+x*y^(n-2)+y^(n-1)=(x^n-y^n)/(x-y)

    ②等比数列求和公式:(q^n-1)a1/(q-1)

    其中q为公比,n为项数,a1为首项