Question

已知数列{an}满足a1=1，an=an-1+n(n≥2)。(1)求a2,a3的值 (2)求数列{an}的通项公式 (3)求数列{1/an}的Tn

Answer 1

a2=a1+1=1+1=2
a3=a2+2=2+2=4

an=a(n-1)+n
an-a(n-1)=n
..............
a3-a2=3
a2-a1=2
以上等式相加得
an-a1=2+3+..........+n
an-a1=(2+n)(n-1)/2
an-1=(2+n)(n-1)/2
an=(2+n)(n-1)/2+1
an=(n^2+n-2+2)/2
an=(n^2+n)/2
an=n(n+1)/2

1/an=2/n(n+1)
1/a1=2/1*2

Tn=2/1*2+2/2*3+...........+2/n(n+1)
=2[1/1*2+1/2*3+.............+1/n(n+1)]
=2*[1-1/2+1/2-1/3+...........+1/n-1/(n+1)]
=2*[1-1/(n+1)]
=2*n/(n+1)
=2n/(n+1)

Answer 2

(1)a2=a1+2=3
a3=a2+3=6
(2)a1=1
a2-a1=2
a3-a2=3
a4-a3=4
......
an-an-1=n
an=n*(1+n)/2
(3)设cn=1/an
则cn=2/n*(n+1)
Tn=2*(1-1/2+1/2-1/3+...+1/n-1/n+1)
=2*(1-1/n+1)
=2n/(n+1)

不懂可追问。若满意望采纳~ ^_^

Answer 3

a1=1
a2=a1+1=2
a3=a3+2=4
.................
an=an-1+n
以上左右相加后：
Sn=Sn-1+(n^2+n)/2
即an=(n^2+n)/2
数列{1/an}的
Tn=2【1-1/（n+1）】
1/an=2/(n^2+n)=2/【n（n+1）】

Answer 4

an=an-1+n
an-a(n-1)=n
a2=a1+2=3
a3=a2+3=6
a2=3 a3=6.

在an-a(n-1)=n中令n=1,2,3,……n
a2-a1=2
a3-a2=3
a4-a3=4
……
an-a(n-1)=n
将上面n-1个等式相加
an-a1=2+3+4+5+……+n
an=a1+2+3+4+5+……+n
an=1+2+3+4+5+……+n
an=n(n+1)/2

1/an=2/[n(n+1)]
Tn=1/a1+1/a2+1/a3+……+1/an
=2[1/(1*2)+1/(2*3)+1/(3*4)+……+1/[n(n+1)]
=2[1/1-1/2+1/2-1/3+1/3-1/4+……+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)

Answer 5

a2=a1+2=3
a3=a2+3=6
an=(an-1)+n
=((an-2)+(n-1))+n
=...
=a1+2+3+...+n
=1+2+...+n
=n(n+1)/2 (n>=2)
Tn=1/an=2/n(n+1)
Tn:{1,1/3,1/6,...,2/n(n+1)}