已知数列｛An｝的前n项和为Sn=n（2n+1）,(I)求该数列的通项公式.(II)判断An=39是该数列的第几项.

Question

已知数列｛An｝的前n项和为Sn=n（2n+1）,(I)求该数列的通项公式.(II)判断An=39是该数列的第几项.

Answer 1

a1 = S1 = 1*(2*1+1) = 3
an = Sn-S(n-1) = n(2n+1)-(n-1)[2(n-1)+1] = n(2n+1)-(n-1)(2n-1) = 2n^2+n-2n^2+3n-1 = 4n-1
当 n=1 时,a1=3 符合通项公式 an=4n-1
所以 an = 39 = 4n-1
解得 n=10