Question

已知偶函数f（x）＝ax^2+bx经过点(1,1),Sn为数列{an}的前n项和,点(n,Sn)(n∈N*)在曲线y=f

Answer 1

(1)f(x)=ax²+bx

f(-x)=f(x)

ax²-bx=ax²+bx

b=0

f(x)=ax²

1=a

f(x)=x²

(2)点(n,Sn)y=f(x)

Sn=n²

an=Sn-Sn-1=n²-(n-1)²=(n+n-1)(n-n+1)=2n-1

(3)Tn=a1b1+a2b2+...+anbn

an:1 3 5 7...2n-1

bn:3 1 3 5.2n-3

Tn=a1b1+a2b2+...+anbn

=3+1*3+3*5+5*7+...+(2n-1)(2n-3)

(Tn)-3=1*3+3*5+5*7+...+(2n-1)(2n-3)

=(2-1)(2+1)+(4-1)(4+1)+(6-1)(6+1)+...+(2n-2-1)(2n-2+1)

=2²-1+4²-1+6²-1+...+(2n-2)²-1

=2²+4²+6²+...+(2n-2)²-n

=4[1+2²+3²+...+(n-1)²]-n

=4[(n-1)n(2n-1)/6]-n

Tn=4[(n-1)n(2n-1)/6]-n+3