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
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 00:07:23
![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](/uploads/image/z/4810569-33-9.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97an%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E2%88%88%E6%AD%A3%E6%95%B4%E6%95%B0%E9%83%BD%E6%9C%89%EF%BC%88kn%2Bb%EF%BC%89%EF%BC%88a1%2Ban%EF%BC%89%2Bp%3D2%EF%BC%88a1%2Ba2%2B...%2Ban%EF%BC%89%2C%E5%85%B6%E4%B8%ADk%2Cb%2Cp%E4%B8%BA%E5%B8%B8%E6%95%B0.%EF%BC%881%EF%BC%89%E5%BD%93k%3D0%2Cb%3D3%2Cp%3D%EF%B9%A34%E6%97%B6+%E6%B1%82a1%2Ba2%2B...an%EF%BC%882%EF%BC%89%E5%BD%93k%3D1%2Cb%3D0%2Cp%3D0%E6%97%B6%2C%E8%8B%A5a3%3D3%2Ca9%3D15+%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F+%E6%88%91%E6%9C%89%E6%80%A5%E4%BA%8B%E5%95%8A+%E7%AC%AC%E4%B8%80%E4%BD%93%E6%88%91%E4%BC%9A)
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
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
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.