Question

已知数列{a_n}是等差数列，设b_n=a_n+1²-a_n².

(1)求证：数列{b_n}是等差数列；

(2)若a_n=-n+2,c_n=,求数列{c_n}的前n项和S_n.

Answer 1

(1)证明：设数列{a_n}的公差为d，则b_n+1-b_n

=(a_n+2²-a_n+1²)-(a_n+1²-a_n²)

=(a_n+2-a_n+1)(a_n+2+a_n+1)-(a_n+1-a_n)(a_n+1+a_n)

=d(a_n+2+a_n+1-a_n+1-a_n)

=d(a_n+2-a_n)

=2d²为常数.

∴数列{b_n}是等差数列.

(2)解：∵a_n=-n+2,

∴公差d=a_n+1-a_n=-1.

∴数列{b_n}的公差为2.

∴b_n=b₁+2·(n-1)=a₂²-a₁²+2n-2=2n-3.

∴c_n=.

∴S_n=                ①

                         ②

①-②得

=

∴S_n=