1.数列{an}的前n项的和Sn=2n（2的n次方）-1,求a6+a7+a8+a9+a10+的值.2.已知数列{an}满足a1=1,an+1=2an/an+2,写出前5项,归纳其通项公式.

Question

1.数列{an}的前n项的和Sn=2n（2的n次方）-1,求a6+a7+a8+a9+a10+的值.
2.已知数列{an}满足a1=1,an+1=2an/an+2,写出前5项,归纳其通项公式.

Answer 1

1.a6+a7+a8+a9+a10=S10-S5=2^10-2^5=2^5(2^5-1)=32*31=992
2.
a(n+1)=2an/(an + 2)
1/[a(n+1)]= (an + 2)/2an = 1/2 + 1/an
1/[a(n+1)] - 1/an = 1/2,∴1/an 为等差数列,公差是1/2
1/an = 1/1 + 1*(n-1)/2 = (n+1)/2
∴an = 2/(n+1)
∴a1=1,a2=2/3,a3=1/2,a4=2/5,a5=1/3