求解

alex318

11  数列，a_0=3， a_(n+1)=2*a_n -1 求a_20-a_19。 答案是2^19
【v2】a1=3,an+1=2an-1,a20-a19=?选2^19

这一题我怎么觉得应该是2^20才对啊

澹泓幽客

这道我也不会诶。。。等牛人。。。

alex318

我计算过程如下
a_20 - a_19 = 2*a_19 - 1 - a_19
= a_19 - 1
= 2*a_18 - 1 - 1
= 2(a_18 - 1)
= ...
=2^2(a_17 - 1)
= ...
...
=2^19(a_0 - 1)
=2^19(3 - 1)
=2^20

alex318

N人给解答啊。。。。。。

heinz

同问，我算的也是2^20

fangfu

You can see the pattern for the nth term of the series from the following:
a_n=2*a_(n-1) -1
     =4*a_(n-2) - (2+1)
     =8*a_(n-3) - (4+2+1)      

a_n=2^n*a_0 - [2^(n-1)+2^(n-2)+...+2^0)]

Then apply the pattern to the 20th and 19th term.

alex318

You can see the pattern for the nth term of the series from the following:
a_n=2*a_(n-1) -1
     =4*a_(n-2) - (2+1)
     =8*a_(n-3) - (4+2+1)      

a_n=2^n*a_0 - [2^(n-1)+2^(n-2)+...+2^0)]

Then apply the pattern to the 20th and 19th term.

-- by 会员 fangfu (2010/10/22 1:18:44)

OK,  As your method, we can get
a_20 = (2^20)*a_0 - [2^19 + 2^18 + ...               + 2^0]
a_19 = (2^19)*a_0 - [              2^18 + 2^17 + ... + 2^0]
So
a_20 - a_19 = (2^20)*a_0 - 2^19 - (2^19)*a_0
                     =(2^20)*3 - 2^19 - (2^19)*3
                     =(2^20)*3 - (2^19)*(1 + 3)
                     =(2^20)*3 - (2^19)*4
                     =(2^20)*3 - (2^19)*2*2
                     =(2^20)*3 - (2^20)*2
                     =(2^20)


Am I right?

hannan518

我计算过程如下
a_20 - a_19 = 2*a_19 - 1 - a_19 =>应该是=2a_18-a_19吧？
= a_19 - 1
= 2*a_18 - 1 - 1
= 2(a_18 - 1)
= ...
=2^2(a_17 - 1)
= ...
...
=2^19(a_0 - 1)
=2^19(3 - 1)
=2^20

-- by 会员 alex318 (2010/10/20 15:47:31)

sunfengchuan

同解 2^20......这道题出自哪里啊？？