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数学
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在等差数列{an}中,a2=9,a5=21,设bn=2^an,求数列{bn}的前n项和sn
人气:250 ℃ 时间:2019-08-21 23:06:12
解答
等差数列,所以an=a1+(n-1)d
y由a2=9,a5=21,可以根据上面的式子算出a1=5,d=4
所以an=4n+1
所以bn=2^4n+1
bn+1/bn=2^4(n+1)+1/2^(4n+1)=2^4=16
所以bn是首相为32,等比q为16的等比数列
所以Sn=b1(1-q^n)/(1-q)=32(16^n -1)/15
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