已知数列an中,a1=a,an+1=(4n+6)an+4n+10/2n+1,n∈N+

数列an中,a1=a,a=[(4n+6)an+4n+10]/(2n+1),n∈N+,
∴(a+2)/(2n+3)=2(an+2)/(2n+1),
∴{(an+2)/(2n+1)}是等比数列,公比为2.
(2)a1=1,(an+2)/(2n+1)=2^(n-1)*(a1+2)/3=2^(n-1),
∴an=(2n+1)*2^(n-1),
∴Sn=3+5*2+7*2^2+……+(2n+1)*2^(n-1),①
2Sn= 3*2+5*2^2+……+（2n-1)*2^(n-1)+(2n+1)*2^n,②
①-②,-Sn=3+2[2+2^2+……+2^(n-1)]-(2n+1)*2^n
=3+2(2^n-2)-(2n+1)*2^n
=-1-(2n-1)*2^n,
∴Sn=1+(2n-1)*2^n,是递增的.