∵数列{a[n]}满足a[1]+2a[2]+3a[3]+...+na[n]=(n+1)(n+2)∴a[1]+2a[2]+3a[3]+...+na[n]+(n+1)a[n+1]=(n+2)(n+3)将上面两式相减,得：(n+1)a[n+1]=2(n+2)∴a[n+1]=2(n+2)/(n+1) 即：a[n]=2(n+1)/n (n≥2)∵a[1]=(n+1)(...