Question

已知数列{an}的通项公式为an=3n-1,求此数列的前7项的和。

Answer 1

解：（1）数列{an}的前n项的和sn
=
3/[2(3n
–
1)]
；
1）当n
=
1时，a1
=
s1
=
3/[2*(3
–
1)]
=
3/4
；
2）当n
≥
2，n∈n*时，an
=
sn
–
sn-1
=
3/[2(3n
–
1)]
–
3/{2[3(n
–
1)
–
1]}
=
3/[2(3n
–
1)]
–
3/[2(3n
–
4)]
=
(3/2)[1/(3n
–
1)
–
1/(3n
–
4)]
=
(3/2)*[(3n
–
4)
–
(3n
–
1)]/[(3n
–
1)(3n
–
4)]
=
(-9/2)/[(3n
–
1)(3n
–
4)]
=
-9/[2(3n
–
1)(3n
–
4)]
；
当n
=
1时，an
=
-9/[2*2*(-1)]
=
9/4
≠
3/4，所以a1
不符合上式；
综上所述，数列{an}的通项公式为：
an
=
{3/4
，当n
=
1时；
....{-9/[2(3n
–
1)(3n
–
4)]，当n
≥
2，n∈n*时。