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?
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/27 08:48:03
![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?](/uploads/image/z/14839128-0-8.jpg?t=a1%3D2%2F5%2C%E4%B8%942an-2a%EF%BC%88n%2B1%EF%BC%89%3D3an%2Aa%EF%BC%88n%2B1%EF%BC%89a1%3D2%2F5%2C%E4%B8%942an-2a%EF%BC%88n%2B1%EF%BC%89%3D3ana%EF%BC%88n%2B1%EF%BC%89+%E8%AF%95%E9%97%AE%E6%95%B0%E5%88%97an%E4%B8%AD%2C%E4%BA%BA%E5%B7%B2%E8%BF%9E%E7%BB%AD%E4%B8%A4%E9%A1%B9%E7%9A%84%E4%B9%98%E7%A7%AFak%2Aa%EF%BC%88k%2B1%EF%BC%89%E6%98%AF%E5%90%A6%E4%BB%8D%E4%B8%BAan%E4%B8%AD%E7%9A%84%E9%A1%B9%2C%E5%A6%82%E6%9E%9C%E6%98%AF%2C%E6%98%AF%E7%AC%AC%E5%87%A0%E9%A1%B9%3F)
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?
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?
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