已知数列{an}：0,-2,-2,0,4,10,18 求{an}的通项公式及前n项和

这个好说，呵呵，请看下面详细
观察相邻两项之间的差
-2-0 =-2
-2- -2 =0
0--2 =2
4-0 =4
10-4 =6
18-10=8
差构成了一列等差数列，首项为-2，公差为2
即设bn=an+1-an, b1=-2,d=2
所以可以知道bn=-2+2(n-1)=2n-4
而
an=an-1+bn-1=an-2+bn-2+bn-1=a1+b1+b2+...+bn-1
=a1+Tn-1(Tn-1表示前n-1项bn的和)
=0+(n-1)*b1+(n-1)(n-1-1)/2*d
=(n-1)*(-2)+(n-1)(n-2)/2*2
=(n-1)(n-4)
而Sn=从n=1到n，求和(n-1)(n-4)
=sigma n^2-5n+4
=n(n+1)(2n+1)/6-5*n(n+1)/2+4n
=n/6*[2n^2+3n+1-15n-15+24]
=n/6*(2n^2-12n+10)
=n(n-1)(n-5)/3

观察相邻两项之间的差
-2-0 =-2
-2- -2 =0
0--2 =2
4-0 =4
10-4 =6
18-10=8
差构成了一列等差数列,首项为-2,公差为2
即设bn=an+1-an, b1=-2,d=2
所以可以知道bn=-2+2(n-1)=2n-4
而
an=an-1+bn-1=an-2+bn-2+bn-1=a1+b1+b2+...+bn-1
=a1+Tn-1(Tn-1表示前n-1项bn的和)
=0+(n-1)*b1+(n-1)(n-1-1)/2*d
=(n-1)*(-2)+(n-1)(n-2)/2*2
=(n-1)(n-4)
而Sn=从n=1到n,求和(n-1)(n-4)
=sigma n^2-5n+4
=n(n+1)(2n+1)/6-5*n(n+1)/2+4n
=n/6*[2n^2+3n+1-15n-15+24]
=n/6*(2n^2-12n+10)
=n(n-1)(n-5)/3