Question

1.设数列{an}的前n项和为sn,若对于所有的正整数n,都有sn=n(a1+an)\2,证明{an}是等差数列2.已知

Answer 1

1.用数学归纳法证明：

1）当n=3时,S3=3(a1+a3)/2

得a1-2a2+a3=0,命题成立；(想一想,为什么初值定为x=3?)

2）假设当n=k时,命题成立,即a1,a2,...,ak是等差数列

设ak=pk+q,则Sk=p/2*k(k+1)+qk=p/2*k^2+(p/2+q)k

则当n=k+1时,

S(k+1)=(k+1)(a1+a(k+1))/2=Sk+a(k+1)

即a(k+1)=[2Sk-(k+1)a1]/(k-1)=[pk^2+(p+2q)k-(k+1)(p+q)]/(k-1)=[pk^2+qk-(p+q)]/(k-1)

=[p(k+1)(k-1)+q(k-1)]/(k-1)=p(k+1)+q

所以a1,a2,...,a(k+1)也是等差数列

2.没看懂题目,修改一下吧