求和:Sn=a+2a2+3a3+…+nan(a≠0,n∈N*)分析:分a=1,a≠1两种情况求解,当a=1时为等差数列易求;当a≠1时利用错位相减法即可求得。解:(1)当a=1时,;(2)当a≠1时,Sn=a+2a2+3a3+…+nan……①①×a得,aSn= a2+2a3+3a4+……+nan+1……②①-②得,(1-a)Sn=a+(2-1)a2+(3-2)a3+(4-3)a4……+(n-n+1)an-nan+1(1-a)Sn=a+a2+a3+a4+……+an-nan+1=a(1-an)/(1-a) -nan+1∴综上所述,当a=1时,;当a≠1时,.求和Sn=1+3x+5x2+7x3+……..+(2n-1)·xn-1(x≠0,n∈N*)解:当x=1时,Sn=1+3+5+…..+(2n-1)=n2当x≠1时,Sn=1+3x+5x2+7x3+……..+(2n-1)xn-1∴xSn=x+3x2+5x3+7x4……..+(2n-1)xn∴两式相减得:(1-x)Sn=1+2x[1+x+x2+x3+...+xn-2]-(2n-1)xn化简得:求和:解:①①两边同时乘以②①-②得:已知数列{an}中,a1=3,点(an,an+1)在直线y=x+2上。(1)求数列{an}的通项公式;(2)若bn=an`3n,求数列{bn}的前n项和Tn。解:(1)∵点(an,an+1)在直线y=x+2上∴an+1=an+2,即an+1-an=2∴数列{an}是以3为首项,以2为公差的等差数列∴an=3+2(n-1)=2n+1(2)∵bn=an·3n∴bn=(2n+1)·3n∴Tn=3×3+5×32+7×33+…+(2n-1)·3n-1+(2n+1)·3n①3Tn=3×32+5×33+…+(2n-1)·3n+(2n+1)·3n+1②由①-②得-2Tn=3×3+2(32+33+…+3n)-(2n+1)·3n+1=9+2×9(1-3n-1)/(1-3)-(2n+1)·3n+1=-2n·3n+1∴Tn=n·3n+1