解ï¼
(1)
a1=2>0ï¼å设ak>0ï¼(kâN*)ï¼åa(k+1)=â(3+ak²)>0ï¼k为任ææ£æ´æ°ï¼å æ¤å¯¹äºä»»ææ£æ´æ°nï¼anæ>0ï¼æ°åå项å为æ£ã
nâ¥2æ¶ï¼an=â[3+a(n-1)²]
an²=3+a(n-1)²
an²-a(n-1)²=3ï¼ä¸ºå®å¼
a1=2ï¼a1²=4ï¼æ°å{an²}æ¯ä»¥4为é¦é¡¹ï¼3为å
¬å·®ççå·®æ°å
an²=4+3(n-1)=3n+1
an>0ï¼an=â(3n+1)
æ°å{an}çé项å
¬å¼ä¸ºan=â(3n+1)
(2)
a1=3>0ï¼å设ak>0ï¼(kâN*)ï¼åa(k+1)=ak²>0ï¼k为任ææ£æ´æ°ï¼å æ¤å¯¹äºä»»ææ£æ´æ°nï¼anæ>0ï¼æ°åå项å为æ£ã
a(n+1)=an²
log3[a(n+1)]=log3(an²)=2log3(an)
log3[a(n+1)]/log3(an)=2ï¼ä¸ºå®å¼
log3(a1)=log3(3)=1
æ°å{log3(an)}æ¯ä»¥1为é¦é¡¹ï¼2为å
¬æ¯ççæ¯æ°åã
log3(an)=1·2^(n-1)=2^(n-1)
an=3^[2^(n-1)]
æ°å{an}çé项å
¬å¼ä¸ºan=3^[2^(n-1)]
温馨提示:答案为网友推荐,仅供参考