a(n+1)=2a2-3^n,求通项公式an

1个回答

  • (1)

    a(n+1)=2a2-3^n

    n=1,a(2)=2a(2)-3,a(2)=3

    a(n+1)=6-3^n

    a(n)=6-3^(n-1)

    (2)

    n为偶数时,

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

    =1-3+5-7+...+(2n-3)-(2n-1)

    =-2-2+...-2

    =-2*n/2

    =-n

    n为奇数时;

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

    =1-3+5-7+...+(2n-5)-(2n-3)+(2n-1)

    =-2*(n-1)/2+(2n-1)

    =n

    (3)

    11...11{共2n个1}-22.22{共n个2}

    =[10^(2n-1)+...+10+1]-2*[10^(n-1)+...+10+1]

    =[10^(2n)-1]/9-2*(10^n-1)/9

    =[10^(2n)-2*10^n+1]/9

    =(10^n-1)^2/9

    根号(11...11{共2n个1}-22.22{共n个2})

    =(10^n-1)/3

    Sn=根号(11-2)+根号(1111-22)+.+根号(11...11{共2n个1}-22.22{共n个2})

    =(10^1-1)/3+(10^2-1)/3+...+(10^n-1)/3

    =(10^1+10^2+...+10^n)/3-n/3

    =[10^(n+1)-10]/27-n/3

    =[10^(n+1)-10-9n]/27

    (4)

    S(n)=2^n-1

    S(n-1)=2^(n-1)-1

    a(n)=S(n)-S(n-1)=2^(n-1)

    a1^2+a2^2+...+an^2

    =2^0+2^2+...+2^(2n-2)

    =(4^n-1)/3