Question

一道关于数列的数学题,已知数列 bn=3的n-1次方(c1/b1)+(c2/b2)+(c3/b3).+(cn/bn)=2

Answer 1

(c1/b1)+(c2/b2)+(c3/b3).+(cn/bn)=2n+1 (1)

(c1/b1)+(c2/b2)+(c3/b3).+(cn-1/b-14n)+=2(n-1)+1 =2n-1 (2)

(1)-(2)得 cn/bn=2

所以cn=2bn=2*3^(n-1)

c1+c2+c3+...+c2008

=2(b1+b2+b3+...+b2008)

=2(1-3^2008)/(1-3)

=3^2008-1