Question

已知等差数列｛an｝和｛bn｝的前n项和分别为An和Bn,且An/Bn=（7n+41）/（n+3）,

则使得an/bn为整数的正整数

Answer 1

解：An/Bn=（7n+41）/（n+3）
=[n/2*(a1+an)]/[n/2*(b1+bn)]
=(a1+an)/(b1+bn)
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=[2a1/d2+(n-1)d1/d2]/[2b1/d2+n-1]
=[d1/d2*n+2a1/d2-d1/d2]/(n+2b1/d2-1)
故有d1/d2=7
2a1/d2-d1/d2=41
2b1/d2-1=3
也即d1=7d2
a1=24d2
b1=2d2
于是an/bn=[a1+(n-1)d1]/[b1+(n-1)d2]
=[24d2+7(n-1)d2]/[2d2+(n-1)d2]
=(7n+17)/(n+1)
=(7n+7+10)/(n+1)
=7+10/(n+1)
故当n=1时，an/bn=12；
当n=4时，an/bn=9；
当n=9时，an/bn=8
也即只有n=1，4，9时，an/bn才为正整数。