设数列an,对任意n∈正整数都有（kn+b）（a1+an）+p=2（a1+a2+...+an）,其中k,b,p为常数.

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设数列an,对任意n∈正整数都有（kn+b）（a1+an）+p=2（a1+a2+...+an）,其中k,b,p为常数.
（1）当k=0,b=3,p=﹣4时求a1+a2+...an
（2）当k=1,b=0,p=0时,若a3=3,a9=15 求an通项公式我有急事啊
第一体我会

k=0,b=3,p = -4.
3[a(1)+a(n)] - 4 = 2[a(1) + a(2)+...+a(n)],3[a(1)+a(1)] - 4 = 2a(1),4a(1)-4=0,a(1)=1.
3[a(1) + a(n+1)] - 4 = 2[a(1)+a(2) + ...+ a(n)+a(n+1)] = 2[a(1)+a(2)+...+a(n)] + 2a(n+1)
= 3[a(1)+a(n)] - 4 + 2a(n+1).
3a(n+1) = 3a(n) + 2a(n+1),
a(n+1) = 3a(n).
{a(n)}是首项为a(1)=1,公比为3的等比数列.
a(n) =3^(n-1).
2[a(1)+a(2)+...+a(n)] = 3[a(1) + a(n)] - 4 = 3[1 + 3^(n-1)] - 4 = 3^n - 1,
a(1)+a(2)+...+a(n) = [3^n - 1]/2.
k=1,b=0,p=0,a(3) = 3,a(9) = 15.
n[a(1)+a(n)] = 2[a(1)+a(2)+...+a(n)],
(n+1)[a(1)+a(n+1)] = 2[a(1)+a(2)+...+a(n)+a(n+1)] = 2[a(1)+a(2)+...+a(n)] + 2a(n+1)
= n[a(1)+a(n)] + 2a(n+1),
(n+1)a(n+1) + a(1) = na(n) + 2a(n+1),
(n-1)a(n+1) = na(n) - a(1),
na(n+2) = (n+1)a(n+1) - a(1),
a(n+2)/(n+1) = a(n+1)/n - a(1)/[n(n+1)] = a(n+1)/n - a(1)/n + a(1)/(n+1),
[a(n+2)-a(1)]/(n+1) = [a(n+1)-a(1)]/n,
{[a(n+1)-a(1)]/n}是首项为[a(2)-a(1)]/1 = a(2)-a(1),的常数数列.
[a(n+1)-a(1)]/n = a(2)-a(1),
a(n+1) - a(1) = n[a(2)-a(1)],
a(n+1) = a(1) + n[a(2)-a(1)],
a(n) = a(1) + (n-1)[a(2)-a(1)].
3 = a(3) = a(1) + 2[a(2)-a(1)] ,
15 = a(9) = a(1)+ 8[a(2)-a(1)]
12 = 15 - 3 = 6[a(2)-a(1)],a(2)-a(1)=2.
a(1) = 3 - 2[a(2)-a(1)] = 3 - 2*2 = -1.
a(n) = a(1) + (n-1)[a(2)-a(1)] = -1 + 2(n-1) = 2n - 3.

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