Question

这个,好象也是初三的数学题.请把过程也写出来.谢谢啦.

Answer 1

AD1=1/3AB
AD2=AD1+D1D2=1/3*AB+1/3*2/3*AB
同理 AD3=1/3*AB+1/3*2/3*AB+1/3*2/3*2/3AB
则AD(n)=1/3*AB+1/3*2/3*AB+...+1/3*(2/3)^(n-1)*AB

同理D(n)E(n)=1/3*BC+1/3*2/3*BC+...+1/3*(2/3)^(n-1)*BC
=1/3*a+1/3*2/3*a+...+1/3*(2/3)^(n-1)*a
可以看出是等比数列，a1=1/3*a,q=2/3
则S(n)=a1*(1-q^n)/(1-q)=a*(1-(2/3)^n) =a*(3^n-2^n)/3^n

Answer 2

你好

AD1=1/3AB

AD2=AD1+D1D2
=1/3*AB+1/3*2/3*AB

AD3=1/3*AB+1/3*2/3*AB+1/3*2/3*2/3AB

AD(n)=1/3*AB+1/3*2/3*AB+...+1/3*(2/3)^(n-1)*AB

D(n)E(n)=1/3*BC+1/3*2/3*BC+...+1/3*(2/3)^(n-1)*BC
=1/3*a+1/3*2/3*a+...+1/3*(2/3)^(n-1)*a

a1=1/3*a,q=2/3

则S(n)=a1*(1-q^n)/(1-q)=a*(1-(2/3)^n) =a*(3^n-2^n)/3^n