bn=3^2n+5,求{bn}的sn

heg4292022-10-04 11:39:541条回答

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坏习惯成自然 共回答了23个问题 | 采纳率87%
Sn=a1(1-q^n)/(1-q)=9(1-9^n)/(1-9)+5n
1年前

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itlv635 共回答了17个问题 | 采纳率88.2%
(1+1/3)(1+1/3^3)=1+1/3+1/3^3+1/3^4
(1+1/3)(1+1/3^3)(1+1/3^5)=1+1/3+1/3^3+1/3^4+1/3^5+1/3^6+1/3^8+1/3^9
...
(1+1/3+1/3^2+...+1/3^n)x2/2-(1/3^2+1/3^7+...+1/3^(5n+2))x(3^5-1)/(3^5-1)=(2+2/3+2/3^2+...+2/3^n)/2-((3^5-1)/3^2+(3^5-1)/3^7+...+(3^5-1)/3^(5n+2))/(3^5-1)
=3/2-3^5/(3^5-1)=3/2-243/242
lim(3^2n+5^n)/(1+9^n)
lim(3^2n+5^n)/(1+9^n)
3^2n or 5^n
漂泊之泉1年前2
道无 共回答了16个问题 | 采纳率93.8%
除以9^n ,3^2n就是9^n
3 x 3^2n+1-21 x 9^n
3 x 3^2n+1-21 x 9^n
求价
Kevin_R1年前2
飘逸空手 共回答了16个问题 | 采纳率93.8%
3 x 3^2n+1-21 x 9^n
=3 x 3^2n+1-7 x3 x3^2n
=3 x 3^2n+1-7 x3^2n +1
=3^2n +1(3-7)
=-4x 3^2n +1