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.

题目

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

答案

a1^3+a2^3+...+an^3=sn^2a1^3+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...