等差数列{an}的前n项的和sn=pn2+n(n+1)+p+3,则p=______.
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等差数列{an}的前n项的和sn=pn2+n(n+1)+p+3,则p=______.
因为等差数列{an}的前n项的和sn=pn2+n(n+1)+p+3,
所以当n≥2时,an=sn-sn-1
=pn2+n(n+1)+p+3-[p(n-1)2+n(n-1)+p+3]
=(2p+2)n-p,
当n=1时,a1=s1=2p+5,也适合上式,
即2p+5=(2p+2)×1-p,解得p=-3,
故答案为:-3.
所以当n≥2时,an=sn-sn-1
=pn2+n(n+1)+p+3-[p(n-1)2+n(n-1)+p+3]
=(2p+2)n-p,
当n=1时,a1=s1=2p+5,也适合上式,
即2p+5=(2p+2)×1-p,解得p=-3,
故答案为:-3.
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