Question

这数列的通项怎么求

1 -13 -31 -51
an -an-1 =n(n-9)

Answer 1

解：a1=1
a2-a1=2²-9×2
a3-a2=3²-9×3
a4-a3=4²-9×4
···
an-a(n-1)=n²-9×n
左右相加，得
an=(1+2²+3²+···+n²)-(9×2+9×3+9×4+···+9×n)
=(1/6)n(n+1)(2n+1)-(9/2)(n-1)(n+2)

Answer 2

没分？

Answer 3

a1=1
a2=a1+2(2-9)
a3=a2+3(3-9)
......
an=an-1+n(n-9)
相加
an=1+2(2-9)+3(3-9)+...+n(n-9)
=(1+2^2+3^3+4^2+...+n^2)-9(2+3+4+...+n)
=n(n+1)(2n+1)/6-9(n-1)(n+2)/2

Answer 4

an -an-1 =n(n-9)
an-1 - an-2 = (n-1)[(n-1)-9]
.
.
a2-a1=2(2-9
相加：
an-a1=2(2-9)+3(3-9)+...+n(n-9)
an=2(2-9)+3(3-9)+...+n(n-9)+1
an=(1+2*2+3*3+...n*n)-(2+n)*(n-1)*9/2