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.
![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.](/uploads/image/z/5252871-39-1.jpg?t=%E8%8B%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%3D%28%E6%B4%BE%2F12%29%2A%282n%5E2%2Bn%29%28n%E2%88%88N%2A%29%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.)
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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