Question

设无穷等差数列{An}的前n项和为Sn.

Answer 1

A(n)=a+(n-1)d,S(n)=na+n(n-1)d/2.

(1)

A(n)=1.5+n-1=n+0.5,S(n)=1.5n+n(n-1)/2=n[n+2]/2,

S(k^2)=k^2(k^2+2)/2=[S(k)]^2=[k(k+2)/2]^2=k^2[k^2+4k+4]/4,

2k^2+4=k^2+4k+4,k^2=4k,k=4.

(2)

S(k^2)=ak^2+k^2(k^2-1)d/2=[S(k)]^2=[ka+k(k-1)d/2]^2,

a+(k^2-1)d/2=[a+(k-1)d/2]^2=(a-d/2)^2+d^2k^2/4+dk(a-d/2),

0=dk^2[2-d]/4 - dk[a-d/2] + a-d/2-(a-d/2)^2,

0=2-d,d=2.

0=a-d/2,a=d/2=1.

A(n)=1+2(n-1)=2n-1,n=1,2,...

S(n)=n+n(n-1)=n^2.

S(k^2)=(k^2)^2=k^4,

[S(k)]^2=[k^2]^2=k^4=S(k^2).