已知数列｛an｝满足an>0且对一切n属于正整数,都有a1^3+a2^3+...+an^3=sn^2,sn是｛an｝的前n项和.求证：a(n+1)^2-a(n+1)=2sn

已知数列｛an｝满足an>0且对一切n属于正整数,都有a1^3+a2^3+...+an^3=sn^2,sn是｛an｝的前n项和.求证：a(n+1)^2-a(n+1)=2sn
已知数列｛an｝满足an>0且对一切n属于正整数,都有a1^3+a2^3+...+an^3=sn^2,sn是｛an｝的前n项和.
求证：a(n+1)^2-a(n+1)=2sn

已知数列｛an｝满足an>0且对一切n属于正整数,都有a1^3+a2^3+...+an^3=sn^2,sn是｛an｝的前n项和.求证：a(n+1)^2-a(n+1)=2sn
a1^3+a2^3+...+an^3=sn^2
a1^3+a2^3+...+[a(n+1)]^3=[s(n+1)]^2
两式相减得
[a(n+1)]^3=[s(n+1)]^2-sn^2
[a(n+1)]^3=[s(n+1)-sn][s(n+1)+sn]
[a(n+1)]^3=a(n+1)[s(n+1)+sn]
[a(n+1)]^2=s(n+1)+sn
[a(n+1)]^2=sn+a(n+1)+sn
[a(n+1)]^2-a(n+1)=2sn