已知数列{An}的通项An=(n+1)(10/11)^n,试问该数列有没有最大项,若有,求最大项和项数,并求Sn最小值.:∵an + 1 – an = (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11) ∴当n<9时,a n + 1 - an>0即a n +
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![已知数列{An}的通项An=(n+1)(10/11)^n,试问该数列有没有最大项,若有,求最大项和项数,并求Sn最小值.:∵an + 1 – an = (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11) ∴当n<9时,a n + 1 - an>0即a n +](/uploads/image/z/12176339-59-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7BAn%7D%E7%9A%84%E9%80%9A%E9%A1%B9An%3D%28n%2B1%29%2810%2F11%29%5En%2C%E8%AF%95%E9%97%AE%E8%AF%A5%E6%95%B0%E5%88%97%E6%9C%89%E6%B2%A1%E6%9C%89%E6%9C%80%E5%A4%A7%E9%A1%B9%2C%E8%8B%A5%E6%9C%89%2C%E6%B1%82%E6%9C%80%E5%A4%A7%E9%A1%B9%E5%92%8C%E9%A1%B9%E6%95%B0%2C%E5%B9%B6%E6%B1%82Sn%E6%9C%80%E5%B0%8F%E5%80%BC.%EF%BC%9A%E2%88%B5an+%2B+1+%E2%80%93+an+%3D+%28n%2B2%29%28+10%2F11+%29%5En%2B1+%E2%80%93+%28n%2B1%29+%28+10%2F11+%29%5En+%3D+%28+10%2F11+%29%5En%2A%289-n%2F11%29+%E2%88%B4%E5%BD%93n%EF%BC%9C9%E6%97%B6%2Ca+n+%2B+1+-+an%EF%BC%9E0%E5%8D%B3a+n+%2B)
已知数列{An}的通项An=(n+1)(10/11)^n,试问该数列有没有最大项,若有,求最大项和项数,并求Sn最小值.:∵an + 1 – an = (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11) ∴当n<9时,a n + 1 - an>0即a n +
已知数列{An}的通项An=(n+1)(10/11)^n,试问该数列有没有最大项,若有,求最大项和项数,并求Sn最小值.
:∵an + 1 – an = (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11)
∴当n<9时,a n + 1 - an>0即a n + 1 >a n ;
当n=9时a n + 1-a n=0,即a n + 1=an ,
当n>9时,a n + 1- an<0即a n + 1<a n ,
故a1<a2<……<a9 = a10>a11>a12>……,
∴数列{an}中最大项为a9或a10 ,
其值为10•( 10/11)9,其项数为9或10
∵an + 1 – an = (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11) 看这个试子
为什么 (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11)? 是怎么得来的 求详解
已知数列{An}的通项An=(n+1)(10/11)^n,试问该数列有没有最大项,若有,求最大项和项数,并求Sn最小值.:∵an + 1 – an = (n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n = ( 10/11 )^n*(9-n/11) ∴当n<9时,a n + 1 - an>0即a n +
就好比2^5-2^4=2×2^4-2^4=(2-1)×2^4=2^4一样
(n+2)( 10/11 )^n+1 – (n+1) ( 10/11 )^n
把(10/11)^n去掉不管
还剩下(n+2)×10/11-(n+1),这个合并起来就是(9-n)/11