已知an 求Sn(数列求和问题,要求用错位相减法)

3个回答

  • 1.Sn=1×2+2×2^2+3×2^3+.+n×2^n (1)

    2Sn= 1×2^2+2×2^3+.+(n-1)×2^n+n×2^(n+1) (2)

    (2)-(1):Sn=-2-2^2-2^3-.-2^n+n×2^(n+1)

    =-2(1-2^n)/(1-2)+n×2^(n+1)

    =(n-1)×2^(n+1)+2

    2.Sn=1+2×2+3×2^2+.+n×2^(n-1) (3)

    2Sn= 1×2+2×2^2+...+(n-1)×2^(n-1)+n×2^n (4)

    (4)-(3):Sn=-1-2-2^2-.-2^(n-1)+n×2^n

    =-(1-2^n)/(1-2)+n×2^n

    =(n-1)×2^n+1

    3.Sn=1×(1/2)+2×(1/2)^2+3×(1/2)^3+.+n×(1/2)^n (5)

    1/2Sn= 1×(1/2)^2+2×(1/2)^3+.+(n-1)×(1/2)^n+n×(1/2)^(n+1) (6)

    (5)-(6):1/2Sn=1/2+(1/2)^2+(1/2)^3+.+(1/2)^n+n×(1/2)^(n+1)

    =1+(n-2)/2×(1/2)^n

    Sn=(n-2)×(1/2)^n+2

    4.Sn=1×2+3×2^2+5×2^3+.+(2n-1)×2^n (7)

    2Sn= 1×2^2+3×2^3+.+(2n-3)×2^n+(2n-1)×2^(n+1) (8)

    (8)-(7):Sn=-2-2×2^2-2×2^3-.-2×2^n+(2n-1)×2^(n+1)

    =-2-8[1-2^(n-1)]/(1-2)+(2n-1)×2^(n+1)

    =(2n-3)×2^(n+1)+6