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  • Cn=[(n+4)(n+5)]/[(n+1)(n+2)].Sn為數(shù)列{Cn}的前n項(xiàng)和,證明Sn

    Cn=[(n+4)(n+5)]/[(n+1)(n+2)].Sn為數(shù)列{Cn}的前n項(xiàng)和,證明Sn
    數(shù)學(xué)人氣:917 ℃時(shí)間:2019-11-14 01:48:41
    優(yōu)質(zhì)解答
    cn=(n+4)(n+5)/[(n+1)(n+2)]
    = 1 + 6(n+3)/(n+1)(n+2)
    = 1+6[2/(n+1) -1/(n+2) ]
    Sn= c1+c2+..+cn
    = 1+6[1/2 -1/(n+2)] + { summation(i:1->n) 1/(i+1) }
    = 4 -1/(n+2)+ { summation(i:1->n) 1/(i+1) }
    n=1
    L.S=S1=c1 = 30/6=5
    R.S=1+6(1+ln1) =7>L.S
    p(1) is true
    Assume p(k) is true
    ie
    4 -1/(k+2)+ { summation(i:1->k) 1/(i+1) } < k+6(1+lnk)
    for n=k+1
    L.S
    =S(k+1)
    =Sk + c(k+1)
    = 4 -1/(k+2)+ { summation(i:1->k) 1/(i+1) } + (k+5)(k+6)/[(k+2)(k+3)]
    < k+6(1+lnk) + (k+5)(k+6)/[(k+2)(k+3)]
    =k+6(1+lnk) + 1+ (6k+24)/[(k+2)(k+3)]
    =(k+1)+6(1+lnk) +6[2/(k+2) -1/(k+3)]
    =(k+1) +6(1 + [ lnk+2/(k+2) - 1/(k+3)] )
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