a1=2/5,且2an-2a(n+1)=3an*a(n+1)a1=2/5,且2an-2a(n+1)=3ana(n+1) 试问数列an中,人已连续两项的乘积ak*a(k+1)是否仍为an中的项,如果是,是第几项?
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a1=2/5,且2an-2a(n+1)=3an*a(n+1)
a1=2/5,且2an-2a(n+1)=3ana(n+1) 试问数列an中,人已连续两项的乘积ak*a(k+1)是否仍为an中的项,如果是,是第几项?
2an-2a(n+1)=3an*a(n+1)
so 2/a(n+1)-2/an=3
另bn=2/an ,b1=5
{bn}是公差为3的等差数列
bn=5+(n-1)*3=3n+2
so an=2/(3n+2)
ak*a(k+1)=4/[(3k+2)(3k+5)]
如果ak*a(k+1)=an=2/[3n+2]
so 4/[(3k+2)(3k+5)]=2/[3n+2]
so 3n+2=(3k+2)(3k+5)/2=[9k^2+21k+10]/2=3*[3k^2+7k+2]/2+2
so n=3k^2+7k+2
易得an=2/(3n+2)
ak*a(k+1)=4/[(3k+2)(3k+5)]
假设乘积ak*a(k+1)是an中的第m项,则
4/[(3k+2)(3k+5)]=2/(3m+2)
m=(3k+1)(k+2)/2
3k+1与k+2奇偶性相反,m为整数,故假设成立
所以任意连续两项的乘积ak*a(k+1)是an中的第(3k+1)(k+2)/2项
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