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【OG20-P177-206题】
A sequence of numbers a1 ,a2 ,a3 . . . is defined as follows : a1=3 , a2=5 , and every term in the sequence after a2 is the product of all terms in the sequence preceding it, e.g., a3=(a1)(a2) and a4=(a1)(a2)(a3) . If an=t and n>2 , what is the value of an+2 in terms oft?
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分析A选项
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分析B选项
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分析C选项
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分析D选项
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分析E选项
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题目讨论 （6条评论）
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emily666
其实前面数字也是多余啊。
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2020-06-02 20:45:53
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huang642
这种选择题带数字算最简单。。 算a3 a5就行
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2019-05-11 09:36:44
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夜晚半透明
an =a1a2...a(n-1)
然后 an=t 所以 a1a2...a(n-1)=t
所以a（n+1）=t 方
a（n+2）= t²再平方
答案是d
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2019-03-06 20:49:16
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bubbles
term1 = 3 term2 = 5 term3 = (term2)(term1) = (5)(3) = 15 (term2)(term1) term4 = (term3)(term2)(term1) = (15)(5)(3) = 15² term5 = (term4)(term3)(term2)(term1) = (15²)(15)(5)(3) = 15⁴ term6 = (term5)(term4)(term3)(term2)(term1) = (15⁴)(15²)(15)(5)(3) = 15⁸ At this point, we can see the pattern. Continuing, we get.... term7 = 15^16 term8 = 15^32 Each term in the sequence is equal to the SQUARE of term before it If term_n =t and n > 2, what is the value of term_n+2 in terms of t? So, term_n = t term_n+1 = t² term_n+2 = t⁴
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2019-01-04 03:50:10
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EricaKo
an=a1*a2*a3*…*a(n-1)=t,a(n+1)=a1*a2*a3*…*a(n-1)*an=t*an=t^2,a(n+2)=a1*a2*a3*…*a(n-1)*an*a(n+1)=t^2*a(n+1)=t^2*t^2=t^4
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2018-08-03 15:52:04
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LTRLTE
根据题目a4=a3•a3,a5=a1•a2•a3•a4=a3•a3•a3•a3.所以an+2=an四次方=t4
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2017-10-05 00:56:03
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默默回复LTRLTE
带入最简的值进行运算
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2018-05-13 04:26:20
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