Question

数列{an} 的首项为3,{b n}为等差数列且b n =a（n+1)- an (n∈N+),若b3=-2,b10 =1

Answer 1

令bn=b1+(n-1)d

b10-b3=7d=12-(-2)=14,d=2,b1=b3-2d=-2-2*2=-6;

bn=-6+2(n-1)=2n-8

a（n+1)- an =2n-8.1)

an-a(n-1)=2n-10.2）

a(n-1)-a(n-2)=2n-12.3)

.

a2-a1=-6=2n-(6+2n).n)

两边累加,得：a(n+1)-a1=2n^2-n(8+2n+6)/2

即a(n+1)-3=2n^2-7n-n^2=n^2-7n

a(n+1)=n^2-7n+3

an=(n-1)^2-7(n-1)+3=n^2-9n+11