设数列{an}满足a1+2a2+3a3+....+nan=n(n+1)(n+2)
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设数列{an}满足a1+2a2+3a3+....+nan=n(n+1)(n+2)
求通项an
求通项an
解:
令n=1时,a1=1*2*3=6;
依题意:
a1+2a2+3a3+....+nan=n(n+1)(n+2),
a1+2a2+3a3+....+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)
两式相减,得到(n+1)a(n+1)=(n+1)(n+2)(n+3)-n(n+1)(n+2),故a(n+1)=(n+2)(n+3)-n(n+2)=3(n+2),
从而:an=3(n+1) (n≥2)
经检验an(a≥2)也适合a1的情况
故通项an=3(n+1).
a1+2a2+3a3+....+nan=n(n+1)(n+2),a1+2a2+3a3+....+(n-1)a(n-1)=n(n-1)(n+1)
相减得nan=n[(n+1)(n+2)-(n-1)(n+1)]=n(3n+3)
所以an=3(n+1)
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