1.
設(shè)bn=nan,其前n項(xiàng)和為Sn：
a1+2a2+3a3+…+nan=b1+b2+……+bn＝Sn
即Sn=n(n+1)(n+2)
bn=Sn-S(n-1)=n(n+1)(n+2)-n(n-1)(n+1)＝3n(n+1)
an=bn/n=3(n+1)
2.
設(shè)共有2n項(xiàng),公差為d
a2n-a1=(2n-1)d=10
偶數(shù)項(xiàng)和-奇數(shù)項(xiàng)和＝n(a2-a1)=nd=30-24=6
聯(lián)立,得：n=3,2n=6
共有6項(xiàng)
3.
通項(xiàng)：1/(2n-1)(2n+3)=(1/4)·[1/(2n-1)-1/(2n+3)]
原式＝（1/4）·[(1/1-1/5)+(1/3-1/7)+(1/5-1/9)+……+1/(2n-1)-1/(2n+3)]
＝（1/4）·[4/3-1/(2n+1)-1/(2n+3)]
＝1/3-（n+1）/[(2n+1)](2n+3)]
4.
f（n）=1+1/2+1/3+……+1/(3n-1)
f（n+1）=1+1/2+1/3+……+1/(3n-1)+1/(3n)+1/(3n+1)+1/(3n+2)
f(n+1)-f(n)=1/(3n)+1/(3n+1)+1/(3n+2)
.