Question

高一数学数列难题 写出过程 图片的过程不完整 有些是错的

Answer 1

(1)
2Sn=(an)^2+an (1)
n=1

a1^2-a1=0
a1=1
(2)
2S(n-1)=(a(n-1))^2+a(n-1) (2)
(1)-(2)

2an = (an)^2 +an - (a(n-1))^2-a(n-1)
(an)^2 -an - (a(n-1))^2-a(n-1) =0
[an + a(n-1)].[an-a(n-1) -1 ] =0
an-a(n-1) -1=0
an-a(n-1) =1
an -a1 =n-1
an = n
(3)
f(n) = Sn/[(n+50)S(n+1)]
= [n(n+1)/2]/ [ (n+50)(n+1)(n+2)/2 ]
= n/[(n+50)(n+2)]
= (1/48)[1/(n+2) - 1/(n+50)]
max f(n) = f(1) = (1/48)( 1/3 - 1/51) = 1/153

追问

还有后面两题呢

追答

(11)
a1+2a2+...+nan=(n-1)Sn +2n (1)

n=1, a1 = 2
a1+2a2+...+(n-1)a(n-1)=(n-2)S(n-1) +2(n-1) (2)

(1)-(2)
nan = (n-1)Sn -(n-2)S(n-1) +2
n(Sn - S(n-1) )= (n-1)Sn -(n-2)S(n-1) +2
Sn +2 = 2( S(n-1) + 2)
Sn +2 =2^(n-1) .(S1+ 2)
Sn = -2+2^(n+1)
an =Sn-S(n-1) =2^n

p+r= 2q

(ap-1).(ar-1)
= (2^p -1)(2^r-1)
= 2^(p+r) - 2^p- 2^r + 1

(aq-1)^2
= ( 2^q -1)^2
= 2^(2q) - 2^(q+1) + 1
=2^(p+r)- 2^p -2^r +1 = (ap-1).(ar-1)
=>ap-1 , aq-1, ar-1: 等比数列

(12)
Sn= 2an+n^2-3n-2
n=1, a1= 4
Sn =2S(n-1) -n^2+3n+2

let
Sn + k1n^2 + k2n+k3 = 2(S(n-1) + k1(n-1)^2 +k2(n-1) + k3)
n^2系数
k1= -1

n系数
k2 -4k1=3
k2 = -1

常数系数
k3 -2k2 +2k1 = 2
k3 =2
ie
Sn =2S(n-1) -n^2+3n+2
Sn -n^2 -n+2 = 2(S(n-1) -(n-1)^2 -(n-1) + 2)
Sn -n^2 -n+2 = 2^(n-1) . (S1 -1 -1 + 2)
= 2^(n+1)
Sn = n^2+n-2 + 2^(n+1)
an = Sn -S(n-1)
= 2n + 2^n
an-2n = 2^n
{an-2n} 是等比数列

cn = 1/(an-n)
= 1/( n+2^n)
< 1/2^n
Tn =c1+c2+...+cn
= 1/3 +c2+...+cn
< 1/3 + 1/2^2+...+1/2^n
= 1/3 + (1/2)(1- 1/2^(n-1) )
< 5/6