数列{an}中,a1=1,an+1=an+2n-1,且已知an=n^2+pn+q,则pq=?
问题描述:
数列{an}中,a1=1,an+1=an+2n-1,且已知an=n^2+pn+q,则pq=?
答
因为a(n+1)=an+2n-1所以a(n+1)-an=2n-1a2-a1=2*1-1a3-a2=2*2-1a4-a3=2*3-1…an-a(n-1)=2*(n-1)-1上述式子左右相加可得an-a1=2*1+2*2+2*3+…+2*(n-1)-(n-1)an=a1+2*[1+2+3+…+(n-1)]-(n-1)=1-n+1+2*[(n-1)(1+...标答是-4啊..麻烦再看看。a(n+1)=an+2n-1这个我没理解错吧?an=a1+2*[1+2+3+…+(n-1)]-(n-1)=1-n+1+2*[(n-1)(1+n-1)]/2=n²-2n+2 (上面的化简弄错了,这个正确)an=n²+pn+q所以p=-2,q=2因此pq=-2*2=-4