假设{an}是公差为d等差数列,则
sn=a1+(a1+d)+(a1+2d)+....+(a1+(n-1)d)
=na1+d*n(n-1)/2
=0.5d*n^2+(a1-0.5d)n。
反过来设sn=a*n^2+b*n
则an=sn-sn-1=an^2+bn -a(n-1)^2-b(n-1)
=2a*n+b-a
an-an-1=2a,所以是{an}是等差数列
假设{an}是公差为d等差数列,则
sn=a1+(a1+d)+(a1+2d)+....+(a1+(n-1)d)
=na1+d*n(n-1)/2
=0.5d*n^2+(a1-0.5d)n。
反过来设sn=a*n^2+b*n
则an=sn-sn-1=an^2+bn -a(n-1)^2-b(n-1)
=2a*n+b-a
an-an-1=2a,所以是{an}是等差数列