数列｛an｝满足an=1 + an = 4n - 3, 当a1=2时 求前n项和Sn数列｛an｝满足

数列｛an｝满足an=1 + an = 4n - 3, 当a1=2时 求前n项和Sn数列｛an｝满足

a(n+1)+an=4n-3a(n+2)+a(n+1)=4(n+1)-3=4n+1两式相减 a(n+2)-an=4由此可以得到：an的奇数项是a1=2为首相,公差d=4的等差数列；an的偶数项是a2=-1为首相,公差d=4的等差数列.当n为偶数时：奇数项有 n/2个,S奇数=[2a1+(n/2-1)*d]*n/4=n²/2奇数项有 n/2个,S偶数=[2a2+(n/2-1)*d]*n/4=n(n-3)/2Sn=S奇数+S偶数=n²/2+n(n-3)/2=n(2n-3)/2当n为奇数时：奇数项有 (n+1)/2个,S奇数=[2a1+((n+1)/2-1)*d]*(n+1)/4=(n+1)²/2偶数项有 (n-1)/2个,S偶数=[2a2+((n-1)/2-1)*d]*(n-1)/4=(n-4)(n-1)/2Sn=S奇数+S偶数=(n+1)²/2+(n-4)(n-1)/2=(2n²-3n+5)/2======以下答案可供参考======供参考答案1:构造：b(n+1)=a(n+1)+r*（n+1）+s将原式中的a(n)换为b(n),且使得b(n)=-b(n-1),再待定系数即可。供参考答案2:构造：b(n+1)=a(n+1)+r*（n+1）+sa(n+1)+an=4n-3a(n+2)+a(n+1)=4(n+1)-3=4n+1两式相减 a(n+2)-an=4由此可以得到：an的奇数项是a1=2为首相，公差d=4的等差数列；an的偶数项是a2=-1为首相，公差d=4的等差数列。当n为偶数时：奇数项有 n/2个，S奇数=[2a1+(n/2-1)*d]*n/4=n²/2奇数项有 n/2个，S偶数=[2a2+(n/2-1)*d]*n/4=n(n-3)/2Sn=S奇数+S偶数=n²/2+n(n-3)/2=n(2n-3)/2当n为奇数时：奇数项有 (n+1)/2个，S奇数=[2a1+((n+1)/2-1)*d]*(n+1)/4=(n+1)²/2偶数项有 (n-1)/2个，S偶数=[2a2+((n-1)/2-1)*d]*(n-1)/4=(n-4)(n-1)/2Sn=S奇数+S偶数=(n+1)²/2+(n-4)(n-1)/2=(2n²-3n+5)/2

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