若数列{bn}是等差数列,切bn=(2n^2-2n)/(n+c),求非零常数c.

若数列{bn}是等差数列,切bn=(2n^2-2n)/(n+c),求非零常数c.
若数列{bn}是等差数列,切bn=(2n^2-2n)/(n+c),求非零常数c.

若数列{bn}是等差数列,切bn=(2n^2-2n)/(n+c),求非零常数c.
b(n+1)-bn=[2(n+1)^2-2(n+1)]/(n+1+c)-(2n^2-2n)/(n+c)
=[2(n+1)^2-2(n+1)]*(n+c)-(2n^2-2n)(n+c+1)]/(n+c)*(n+c+1)
=2(n^2+2nc+n)/(n+c)*(n+c+1)
=2(n^2+2nc+n)/[n^2+2cn+n+c(c+1)]
因为｛bn}为等差数列,故该结果为常数
所以 c(c+1)=0 c=0或者-1