已知数列an满足a1=1,an=3的n次方+2an-1

2个回答

  • n≥2时,

    an=3ⁿ+2a(n-1)

    等式两边同除以2ⁿ

    an/2ⁿ=(3/2)ⁿ +a(n-1)/2^(n-1)

    an/2ⁿ -a(n-1)/2^(n-1)=(3/2)ⁿ

    a(n-1)/2^(n-1) -a(n-2)/2^(n-2)=(3/2)^(n-1)

    …………

    a2/2²-a1/2=(3/2)²

    累加

    an/2ⁿ- a1/2=(3/2)²+(3/2)³+...+(3/2)ⁿ

    an/2ⁿ=a1/2 +(3/2)²+(3/2)³+...+(3/2)ⁿ

    =1/2+(3/2)²+(3/2)³+...+(3/2)ⁿ

    =(3/2)+(3/2)²+(3/2)³+...+(3/2)ⁿ -1

    =(3/2)×[(3/2)ⁿ-1]/(3/2-1) -1

    =3^(n+1)/2ⁿ -4

    an=3^(n+1) -4×2ⁿ=3^(n+1)-2^(n+2)

    n=1时,a1=3²-2³=1,同样满足通项公式

    数列{an}的通项公式为an=3^(n+1) -2^(n+2)