Question

等差数列{an}的公差为2，其前n项和Sn=pn平方+2n，求p的值及an

Answer 1

Sn=(a1+a1+2(n-1))n/2=n^2+(a1-1)n
所以p=1，a1=3，
an=a1+2（n-1）=2n+1

Answer 2

d=a2-a1=2
a2=S2-S1
a1=S1
则S2-2S1=2
(4p+4)-2(p+2)=2
p=1

a1=S1=1+2=3
d=2
an=2n+1

Answer 3

s2-s1=a2=a1+2
4p+4-p-2=a1+2
3p=a1
又s1=a1
p+2=a1
得p=1，a1=3
an=a1+(n-1)*2=2n+1

Answer 4

Sn=(a1+an)*n/2=[a1+a1+(n-1)*2]*n/2=(a1+n-1)*n=n^2+(a1-1)n=pn^2+2n
比较系数有p=1,a1=3
所以an=a1+(n-1)*2=2n+1

Answer 5

P值是1，an=2n+1.公差d=2因爲Sn=pn平方+2n所以S1=p+2=a1①
S2=4p+4=a1+a2②
a2=a1+2③
.把①代③得a2=p+4④把①和④代入②得4p+4=p+2+p+4
所以p=1
a1=p+2=3
所以an=a1+（n－1）d=2n+1