a1=1,A(n+1)=2^n+2an 求an 通项公式Andy983020126 怎么看出来两边要除以2^(n+1)
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a1=1,A(n+1)=2^n+2an 求an 通项公式Andy983020126 怎么看出来两边要除以2^(n+1)
a1=1,A(n+1)=2^n+2an 求an 通项公式
Andy983020126 怎么看出来两边要除以2^(n+1)
a1=1,A(n+1)=2^n+2an 求an 通项公式Andy983020126 怎么看出来两边要除以2^(n+1)
a(n+1)=2^n+2an (两边同除以2^(n+1) )
a(n+1)/2^(n+1)=an/2^n+1/2
a(n+1)/2^(n+1)-an/2^n=1/2
所以{an/2^n}是以a1/2=1/2为首相,d=1/2为公差的等差数列
an/2^n=1/2-(1/2)(n-1)=-n/2+1
所以an=2^n(-n/2+1)
a(n+1)/2^(n+1)=1/2+an/2^n设bn=a(n+1)/2^(n+1) 则bn=为等差数列 bn=n/2 an=1
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