å¨æ°å{an}ï¼{bn}ä¸ï¼a1=2ï¼a(n+1)-an=6n+2ï¼ç¹(an/n,bn)å¨y=x^3+mxçå¾åä¸ï¼{bn}çæå°é¡¹ä¸ºb3.
ï¼1ï¼æ±{an}çé项å
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ï¼2ï¼æ±mçåå¼èå´ï¼
ï¼1ï¼è§£æï¼âµa(n+1)-an=6n+2ï¼a1=2ï¼
â´a2-a1=8ï¼
a3-a2=14ï¼
a4-a3=20ï¼
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an-a(n-1)=6n-4
两边ç¸å ï¼
â´an-a1=(n-1)(8+6n-4)/2=3n^2-n-2==>an=3n^2-nï¼
ï¼2ï¼è§£æï¼âµ(an/nï¼bn)å¨y=x^3+mxçå¾è±¡ä¸ï¼
â´bn=(3n-1)^3+m(3n-1)
â´b1=8+2mï¼b2=125+5mï¼b3=512+8mï¼b4=1331+11m
âµ{bn}çæå°å¼ä¸ºb3ï¼
â´125+5m>=512+8m==>m<=-129
1331+11m>=512+8m==>m>=-273
â´-273â¤mâ¤-129
b(n+1)=(3n+2)^3+m(3n+2)
â´b(n+1)-bn=(3n+2)^3-(3n-1)^2+m(3n+2)-m(3n-1)
=(3n+2)^3-(3n-1)^3+3m
=3[(3n+2)^2+(3n-1)^2+(3n+2)(3n-1)]+3m
=3(27n^2+9n+3+m)
â´nâ¥3æ¶ï¼b(n+1)-bnï¼0ï¼â´b(n+1)ï¼bnï¼å³æ°åä»ç¬¬3项起æ¯éå¢çï¼
综ä¸å¯å¾ï¼-273â¤mâ¤-129ï¼
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