设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)求通项an
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设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)求通项an设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)求通项an设数列｛an｝满足a1+2a2+
设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)求通项an
设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)
求通项an
设数列｛an｝满足a1+2a2+3a3+.+nan=n(n+1)(n+2)求通项an
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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