若数列{an}的前n项和Sn=(派/12)*(2n^2+n)(n∈N*)，证明：数列{an}是等差数列。

显示全部楼层 · 2011-10-15 23:34:03

证明：Sn=(π/12)*(2n^2+n)
=(π/6)*(n^2)+(π/12)*n
当n≥2时，
S(n-1)=(π/6)*[(n-1)^2]+(π/12)*(n-1)
=(π/6)*(n^2)+(π/12)*n-(π/3)*n+π/12
S(n-2)= (π/6)*[(n-2)^2]+(π/12)*(n-2)
=(π/6)*(n^2)+(π/12)*n-(2π/3)*n+π/2
∴an=Sn-S(n-1)
=(π/3)*n-π/12...

千问 · 2011-10-15 23:34:03

Sn=(π/12)*(2n^2+n)S(n-1)=(π/12)*(2n(n-1)^2+(n-1))Sn-S(n-1)=(π/12)(4n-1)=anS1=(π/12)*3=π/4=a1an=π/4+(n-1)*(π/3)an=a1+(n-1)*(π/3)则d=π/3数列{an}是等差数列得证...