1
∵√Sn - √Sn-1 = √2
∴{√Sn}为等差数列,公差为√2
∵a1=2∴√S1=√2
∴√Sn=√2*n
∴Sn=2n²
2
n≥2时,an=Sn-S(n-1)=2n²-2(n-1)²=4n-2
n=1时,上式也成立
∴数列{an}的通项公式为
an=4n-2
3
bn=1/4*ana(n-1)有问题呀
感觉应该是
bn=1/4*ana(n+1)
=1/4*(4n-2)(4n+2)
=(2n-1)(2n+1)
1/bn=1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
∴1/b1+1/b2+.+1/bn
=1/2[1-1/3+1/3-1/5+1/5-1/7+.+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=1/2-1/(4n+2)
∵1/(4n+2)>0
∴1/2-1/(4n+2)