2sn=nan+n
2s2=2*3+2=8
s2=4
s2=a1+a2
4=a1+3
a1=1
2s3=3*a3+3
2(s2+a3)=3a3+3
2(4+a3)=3a3+3
8+2a3=3a3+3
a3=5
an=2n-1
用數(shù)學(xué)歸納法證明
n=1,成立
若n=k時(shí),Sk=(1/4)(ak+1)^2,ak=2k-1
Sk=(1/4)*(2k-1+1)^2=k^2,
則n=k+1時(shí)
S(k+1)=a(k+1)+Sk=(1/4)[(a(k+1)+1]^2
4a(k+1)+4k^2=[a(k+1)+1]^2
[a(k+1)]^2+2a(k+1)+1-4a(k+1)-4k^2=0
[a(k+1)]^2-2a(k+1)+1-4k^2=0
[a(k+1)-(1+2k)][a(k+1)-(1-2k)]=0
所以a(k+1)=1+2k,a(k+1)=1-2k
因?yàn)閍n是正數(shù)數(shù)列
所以a(k+1)=2k+1=2(k+1)-1
得證
an=2n-1