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.
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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.
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
a(n-1)=S(n-1)-S(n-2)
=(π/3)*n-(5π/12)
∴an-a(n-1)=π/3(为常数)
当n=1时,
S1=π/4,a1=(π/3)*1-π/12=π/4
=>S1=a1
综上所得,数列{an}是等差数列.
=(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
a(n-1)=S(n-1)-S(n-2)
=(π/3)*n-(5π/12)
∴an-a(n-1)=π/3(为常数)
当n=1时,
S1=π/4,a1=(π/3)*1-π/12=π/4
=>S1=a1
综上所得,数列{an}是等差数列.
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