Question

等差数列{an}的公差为2，其前n项和Sn=pn的平方+2n(n属于N)，求p的值及a

求p的值及an

Answer 1

Sn=pn^2+2n
Sn-1=p(n-1)^2+2(n-1)
则an=2pn-p+2
an-1=2p(n-1)-p+2
则d=2p=2 所以p=1
an=2n+1

Answer 2

a1=S1=p+2
Sn=na1+1/2n(n-1)d=n(p+2)+n(n-1)*1/2*2=n(p+2)+n^2-n=n^2+(p+1)n
又Sn=pn^2+2n
故pn^2+2n=n^2+(p+1)n
比较得p=1
a1=p+2=3
an=a1+(n-1)d=3+2(n-1)=2n+1

Answer 3

Sn=a*n+[n(n-1)d]/2
整理后得：Sn=d/2*n^2+(a-d/2)*n
已知d=2,则Sn=n^2+(a-1)*n
又已知Sn=p*n^2+2n
则p=1 a=3

Answer 4

p=1/2 a在哪儿呢？