构造新数列求通项公式VIP免费

构造新数列求通项公式n+1na=qa+f(n)构造新数列法又称待定系数法适用于基本思路是转化为等差数列或等比数列，而数列的本质是一个函数，其定义域是自然数集的一个函数。通过恰当的恒等变形,如配方、因式分解、取对数、取倒数等,转化为等比数列或等差数列.(1)若an+1=pan+q,则:an+1-=p(an-).(3)若an+1=pan+q(n),则:(2)若an+1=,则:panr+qanan+11an1=·+.prpq(4)若an+1=panq,则:lgan+1=qlgan+lgp.例.已知数列{an}满足:a1=1,an+1=2an+3×2n-1,求{an}的通项公式.an=(3n-1)×2n-2an+1pn+1anpn=+.q(n)pn+1例.已知数列{an}中,a1=1,an+1=an+1(nN*),求an.12解法一∵an+1=an+1(nN*),12∴an=an-1+1,an-1=an-2+1.1212两式相减得:an-an-1=(an-1-an-2)12∴{an-an-1}是以a2-a1=为首项,公比为的等比数列.1212∴an-an-1=()n-2=()n-1.121212∴an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1++()2+…+()n-1121212=2-21-n.即an=2-21-n.解法二由解法一知an-an-1=21-n,又an=an-1+1,12消去an-1得an=2-21-n.解法三∵an=an-1+1,12令an+=(an-1+),12则=-2.∴an-2=(an-1-2).12∴{an-2}是以a1-2=-1为首项,公比为的等比数列.1212∴an-2=-()n-1.即an=2-21-n.3.已知数列{an}中,a1=1,an+1=an+1(nN*),求an.12