1 设Sn=1+2+3â¦â¦+nï¼åf(n)=Sn/((n+7)*S(n+1))çæ大å¼ä¸º
sn=(n+1)n/2;
s(n+1)=(n+2)(n+1)/2;
f(n)=n/[(n+7)(n+2)]=1/[9+n+14/n],
æ以å½n=4ï¼ææ大å¼=2/33.
2 设f(x)æ¯ä¸æ¬¡å½æ°ï¼è¥f(0)=1ï¼ä¸f(n)ï¼f(4)ï¼f(13)æçæ¯æ°åï¼åf(2)+f(4)â¦â¦f(2n)çäº
设y=kx+bï¼æ ¹æ®é¢æï¼f(0)=1,æ以b=1.
f(n)ï¼f(4)ï¼f(13)æçæ¯æ°åæï¼
f(4)^2=f(n)f(13)
(4k+1)^2=(4n+1)(13k+1)
n=(16k^2-5k)/(52k-4) ?
3 ä¸çå¼ï¼ax)/(x-1)<1ç解é为x<1æx>2,é£ä¹açå¼çäº
解é为ï¼x<1æx>2
åæ¹ç¨å¯è®¾ä¸ºï¼
(x-1)(x-2)>0....(1)
åä¸çå¼å形为ï¼
ax/(x-1)<1
ax/(x-1)-1<0
(ax-x+1)/(x-1)<0
[(a-1)x+1]/(x-1)<0
[(1-a)x-1]/(x-1)>0
(1-a)[x-1/(1-a)]/(x-1)>0...(2)
æ¯è¾ï¼1ï¼ãï¼2ï¼å¾å°ï¼
1/(1-a)=2,æ以a=1/2.
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