求解两道高一数学题

\u6c42\u89e3\u4e24\u9053\u9ad8\u4e00\u6570\u5b66\u9898

\u4e00\u89e3\uff1a(1)\u2235\u5728\u6570\u5217{a[n]}\u4e2d\uff0ca[n+1]=(1+1/n)a[n]+(n+1)/2^n
\u2234\u4e24\u8fb9\u9664\u4ee5(n+1)\uff0c\u5e76\u52a0\u4e0a2/2^(n+1)\uff0c\u5f97\uff1a
a[n+1]/(n+1)+2/2^(n+1)=a[n]/n+2/2^n
\u2235a[1]=1
\u2234{a[n]/n+2/2^n}\u662f\u5e38\u6570\u4e3aa[1]/1+2/2^1=2\u7684\u5e38\u6570\u6570\u5217
\u5373\uff1aa[n]/n+2/2^n=2
\u2234a[n]/n=2-2/2^n=2(1-1/2^n)
\u2235b[n]=a[n]/n
\u2234b[n]=2(1-1/2^n)

(2)\u2235a[n]=2(n-n/2^n)
\u2234S[n]=2[(1+2+3+...+n)-(1/2^1+2/2^2+3/2^3+...+n/2^n)]

\u4ee4\uff1aT[n]=1/2^1+2/2^2+3/2^3+...+n/2^n
\u5219\uff1aT[n]/2=1/2^2+2/2^3+3/2^4+...+n/2^(n+1)
\u5c06\u4e0a\u9762\u4e24\u5f0f\u76f8\u51cf\uff0c\u5f97\uff1a
T[n]/2=(1/2^1+1/2^2+1/2^3+...+1/2^n)-n/2^(n+1)
=(1/2)[1-(1/2)^n]/(1-1/2)-n/2^(n+1)
=1-(1/2)^n-n/2^(n+1)
=1-(n+2)(1/2)^(n+1)
\u2234T[n]=2-(n+2)(1/2)^n

\u2234S[n]=2{n(n+1)/2-[2-(n+2)(1/2)^n]}
=n(n+1)+(n+2)(1/2)^(n-1)-4


\u4e8c\u89e3\uff1a(1)\u2235\u6570\u5217{2a[n+1]-a[n]}\u662f\u516c\u6bd4\u4e3a1/3\u7684\u7b49\u6bd4\u6570\u5217
\u2234(2a[n+1]-a[n])/(2a[n]-a[n-1])=1/3
\u6574\u7406\uff0c\u5f97\uff1a6a[n+1]=5a[n]-a[n-1]
\u5373\uff1a2a[n+1]-a[n]=(2a[n]-a[n-1])/3 \u30101\u3011
\u2235a[1]=1/3\uff0ca[2]=5/18
\u2234{2a[n+1]-a[n]}\u662f\u9996\u9879\u4e3a2a[2]-a[1]=2/9\uff0c\u516c\u6bd4\u4e3a1/3\u7684\u7b49\u6bd4\u6570\u5217
\u5373\uff1a2a[n+1]-a[n]=(2/9)(1/3)^(n-1)=2(1/3)^(n+1)

\u2234a[n+1]+2(1/3)^(n+1)=[a[n]+2(1/3)^n]/2 \u30102\u3011
\u2234{a[n]+2(1/3)^n}\u662f\u9996\u9879\u4e3aa[1]+2(1/3)^1=1\uff0c\u516c\u6bd4\u4e3a1/2\u7684\u7b49\u6bd4\u6570\u5217
\u5373\uff1aa[n]+2(1/3)^n=(1/2)^(n-1)
\u2234a[n]=(1/2)^(n-1)-2(1/3)^n

(2)S[n]
=[1-(1/2)^n]/(1-1/2)-2(1/3)[1-(1/3)^n]/(1-1/3)
=2[1-(1/2)^n]-[1-(1/3)^n]
=2-(1/2)^(n-1)-1+(1/3)^n
=1-(1/2)^(n-1)+(1/3)^n

\u3010\u6b64\u9898\u7ed9\u51fa\u7684\u4e24\u4e2a\u6761\u4ef6\uff1a{Log[2](3a[n+1]-a[n])}\u662f\u516c\u5dee\u4e3a-1\u7684\u7b49\u5dee\u6570\u5217\uff0c\u4ee5\u53ca\u6570\u5217{2a[n+1]-a[n]}\u662f\u516c\u6bd4\u4e3a1/3\u7684\u7b49\u6bd4\u6570\u5217\uff0c\u5176\u5b9e\u662f\u7b49\u4ef7\u7684\u3002 \u30101\u3011\u3001\u30102\u3011\u5f0f\u53ef\u4ee5\u7528\u5f85\u5b9a\u7cfb\u6570\u6cd5\u83b7\u5f97\uff0c\u4f46\u7531\u4e8e\u90fd\u6bd4\u8f83\u7b80\u5355\uff0c\u76f4\u63a5\u89c2\u5bdf\u5373\u53ef\u83b7\u5f97\u3002\u3011

\u89e3\uff1a\u4f9d\u9898\u610f\uff0c\u51fd\u6570g\uff08x\uff09=loga\uff08a2x+t\uff09\uff08a\uff1e0\uff0ca\u22601\uff09\u5728\u5b9a\u4e49\u57df\u4e0a\u4e3a\u5355\u8c03\u9012\u589e\u51fd\u6570\uff0c\u4e14t\u22650\uff0c
\u800ct=0\u65f6\uff0cg\uff08x\uff09=2x\u4e0d\u6ee1\u8db3\u6761\u4ef6\u2461\uff0c
\u2234t\uff1e0\uff0e\u8bbe\u5b58\u5728[m\uff0cn]\uff0c\u4f7f\u5f97g\uff08x\uff09\u5728[m\uff0cn]\u4e0a\u7684\u503c\u57df\u4e3a[m\uff0cn]\uff0c
\u2234 \uff0c\u5373 \uff0c
\u2234m\uff0cn\u662f\u65b9\u7a0b\uff08ax\uff092-ax+t=0\u7684\u4e24\u4e2a\u4e0d\u7b49\u5b9e\u6839\uff0c
\u2234\u25b3=1-4t\uff1e0\uff0c
\u2234 \uff0c
\u6545\u9009D\uff0e

1.
设f(x)=ax²＋bx＋c,则f(x)＞-2x可化为ax²＋（b＋2）x＋c＞0
由题意得方程ax²＋（b＋2）x＋c＝0的两根是1和3
所以1＋3＝－（b＋2）／a,1×3＝c／a
即b＝-4a－2,c＝3a
f(x)=ax²－（4a＋2）x＋3a
(1)方程f(x)+6a=0可化为ax²－（4a＋2）x＋9a＝0
Δ＝（4a＋2）²－36a²＝0
a＝1,b=-6,c=3或a=-1/5,b=-6/5,c=-3/5
(2)由题意知a<0
最大值为[12a²－（4a＋2）²]／4a＞0
解得a＜－2－√3或a＞－2＋√3
2.
(1)令x=y=1,则f(1)=f(1)-f(1)=0
(2)设x1<x2∈（0,＋∞）,则x2／x1＞1
f(x2)-f(x1)=f(x2/x1)>0
所以f(x)在（0，＋∞）上递增
（3）
f(x+3)-f(1/x)=f[x(x+3)]
所以原不等式可以化为f[x²＋3x]＜f（6）
因此x＋3＞0
1/x＞0
x²＋3x＜6
∴0＜x＜（－3＋√33）／2

如图

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