数列{an}满足a(n+1)+an=4n-3(n∈N*),若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1

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数列{an}满足a(n+1)+an=4n-3(n∈N*),若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1数列{an}满足a(n+1)+an=4n-3(n∈N*),若{an}满足a1=2

数列{an}满足a(n+1)+an=4n-3(n∈N*),若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1
数列{an}满足a(n+1)+an=4n-3(n∈N*),若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1

数列{an}满足a(n+1)+an=4n-3(n∈N*),若{an}满足a1=2,Sn为{an}的前n项和,求S2n+1
因为a(n+1)+an=4n-3
所以S(2n+1)=a1+a2+a3+a4+a5+...+a(2n)+a(2n+1)
=a1+(a2+a3)+(a4+a5)+...+[a(2n)+a(2n+1)]
=2+(4*2-3)+(4*4-3)+...+(4*2n-3)
=2+[4*2*(1+2+...+n)]-3n
=2+4n(n+1)-3n
=4n^2+n+2

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