数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和

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数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和数列〔an〕满足an+1+an=4n-3,当

数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和
数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和

数列〔an〕满足an+1+an=4n-3,当a1=2时,求数列〔an〕前n项和
a(n+1)+an=4n-3,an+a(n-1)=4(n-1)-3故a(n+1)-a(n-1)=4,n≥2 a1=2,a2=-1
n为奇数时an=2+(n-1)/2*4=2n,a(n-1)=-1+(n-1)/2*4=2n-5
Sn=(2+2n)*(n+1)/2/2+(-1+2n-5)*(n-1)/2/2 =n^2-n+2
n为偶数时Sn=n^2+n/2

题目打错了吧

a(n+1)+an=4n-3 an-a(n-1)=4(n-1)-3 两式相减,得到 a(n+1)-a(n-1)=4 也就是说a1,a3,a5,a(2k-1)成等差数列, a2,a4,

an=(-1)^(n-1)x5/2+2n-5/2