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  • 數(shù)列an的前n項(xiàng)和為Sn,已知a1=1,2Sn/n=a(n+1)-n^2/3-n-2/3 ,n屬于正整數(shù)

    數(shù)列an的前n項(xiàng)和為Sn,已知a1=1,2Sn/n=a(n+1)-n^2/3-n-2/3 ,n屬于正整數(shù)

    求a2的值

    2.數(shù)列{an}的通項(xiàng)公式
     
    數(shù)學(xué)人氣:629 ℃時(shí)間:2019-08-20 12:44:56
    優(yōu)質(zhì)解答
    2Sn/n=a(n+1)-n^2/3 -n -2/3
    6Sn = 3n(S(n+1) -Sn) - n^3 - 3n^2 - 2n
    3nS(n+1) = 3(n+2)Sn +n^3 +3n^2 +2n
    =3(n+2)Sn +n(n+1)(n+2)
    3S(n+1)/[(n+1)(n+2)] = 3Sn/[n(n+1)] +1
    S(n+1)/[(n+1)(n+2)] -Sn/[n(n+1)] =1/3
    Sn/[n(n+1)] - S1/[1(1+1)] = (n-1)/3
    Sn/[n(n+1)] = (2n+1)/6
    Sn = (2n+1)n(n+1)/6
    an = Sn -S(n-1)
    = (1/6)[(2n+1)n(n+1) - (2n-1)(n-1)n ]
    = n^2
    a2 = 2^2= 4
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