jj数学

qinyitian

MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-bidi-font-family: PMingLiU;">1.      MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-fareast-font-family: 宋体; mso-bidi-font-family: Tahoma; mso-fareast-language: ZH-CN;">If the sequence x₁, x_2, x_{3, …,} x_n, … is such that x₁  = 3 and x_n+1
                = 2x_n – 1 for n
            ≥MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-fareast-font-family: 宋体; mso-bidi-font-family: Tahoma; mso-fareast-language: ZH-CN;"> 1,MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-bidi-font-family: Tahoma;">
                

MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-fareast-font-family: 宋体; mso-bidi-font-family: Tahoma; mso-fareast-language: ZH-CN;">then x₂₀ – x₁₉ =

MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-fareast-font-family: 宋体; mso-bidi-font-family: Tahoma; mso-fareast-language: ZH-CN;">大家算这个得多少，我算的答案跟写jj的人给的不一样，人家得 2^19MingLiU","serif"; mso-bidi-font-size: 12.0pt; mso-font-kerning: 0pt; mso-bidi-font-family: Tahoma;">

xuan37

应该是2^19

qinyitian

哦，算出来了，刚算错了，谢谢啦！

YPLIN

can you please explain how you get the answer? thanks

yunlinli

X1=3


X2=2(3)-1=5


X3=2(5)-1=9


X4=2(9)-1=17


        

That took like 20 seconds to do quickly and I can see from these
sequences and, **by quickly surveying the answer choices** that there
is something going on with powers of 2. I notice that the increases
from Xn to Xn+1 move by a power of two related to Xn. The sequence can
be written as:


        

Xn = Xn-1 + 2^(n-1)  (you can quickly check that that works as well starting at X1)


        

So, to solve the problem X20 - X19, I'll use that formula to solve


        

X20 = X19 + 2^19 and then subtract X19 without having to expand it gives me:


        

ANS = X19 + 2^19 - X19 = 2^19 [A]

qinyitian

a20-a19=a19-1=2a18-1-1=2(2a17-1)-2=2^2a17-2=……=2^18a1-2^18

a1=3带入就等于2^19了