设数列的前n项的和为sn,a1=2,根号sn-根号sn-1=根号2,求sn还要求an

Question

设数列的前n项的和为sn,a1=2,根号sn-根号sn-1=根号2,求sn
还要求an

Answer 1

√Sn-√S(n-1)=√2
令bn=√Sn
则bn是以√2位公差的等差数列
bn=b1+(n-1)√2
S1=a1=2
所以b1=√S1=√2
所以bn=√2+(n-1)√2=n*√2
所以Sn=(bn)^2=2n^2
an=Sn-S(n-1)=2n^2-2(n-1)^2=2(2n-1),(n>=2)