Question

已知数列{an}满足递推式：an-a(n-1)=2n-1(n≥2,n∈N*),且a1=1,求bn=[（-1）^n]·an

Answer 1

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

.

a3-a2=2*3-1

a2-a1=2*2-1

以上等式相加得

an-a1=2*2-1+2*3-1+.+2n-1

an-1=2*(2+3+.+n)-(n-1)

an-1=2*(n-1)(n+2)/2-n+1

an=n^2+n-2-n+2

an=n^2

b1=[(-1)^1]*a1

=-a1

=-1

b1=[(-1)^n]*an

=(-1)^n*n^2

n为偶数时

Sn=-1^2+2^2+.+n^2

=(2-1)(2+1)+.+(n+n-1)(n-n+1)

=1+2-.+n-1+n

=n(n+1)/2

n为奇数时

Sn=-1^2+2^2+.-n^2

=(2-1)(2+1)+.+(n-1+n-2)(n-1+n-2)-n^2

=n(n-1)/2-n^2

=-n(n+1)/2