Question

高一数学数列，急求详细过程，谢谢

已知数列{an}的前n项的和Sn=n^2+5n-1求通项公式

在等差数列{an}中，已知d=2,an=11 Sn=35，求an和n

Answer 1

1.已知数列{an}的前n项的和Sn=n^2+5n-1求通项公式
an=Sn-Sn-1=(n²+5n-1)-[(n-1)²+5(n-1)-1],化解得到an=2n+4.代n=1验算S1=5,而a1=6，那么就用分段的形式书写即可

Answer 2

1.
Sn=n²+5n-1
an=(Sn-S(n-1)=(n²+5n-1) - [(n-1)²+5(n-1)-1]=2n+4 (n≥2)
当n=1时，a1=S1=5，不满足上式。
∴an= 5 n＝1
2n+4 n≥2

2.
是求a1吧？
Sn=n·a1 + n(n-1)d/2
an=a1+(n-1)d

na1+(n-1)n=35
a1+2(n-1)=11

解得n=7，a1=-1或者n=5，a1=3

Answer 3

通项不是S(n)-S(n-1)么？不就是n^2+5n-1 - (n-1)^2 -5(n-1)+1 = 2n+4

至于等差数列,带入求和公式不久可以了？
d=2, an=11, an = a1 + (n-1)*d = a1 + 2n-2
a1 = an - 2n +2 = 11+2-2n = 13-2n
S(n) = (a1+an)*n/2 = (13-2n + 11) *n /2 = (24-2n)*n /2 = 35
n^2 - 12n +35=0
n=5或者7

Answer 4

a1=s1=1^2+5-1=5

sn=n^2+5n-1
s(n-1)=(n-1)^2+5(n-1)-1
=n^2-2n+1+5n-5-1
=n^2+3n-5
an=sn-s(n-1)
=n^2+5n-1-(n^2+3n-5)
=n^2+5n-1-n^2-3n+5
=2n+4
an=2n+4(n>=2)

an=a1+(n-1)d
11=a1+2(n-1)
a1+2n-2=11
a1=13-2n

sn=(a1+an)*n/2
(a1+11)*n/2=35
(a1+11)*n=70
(13-2n+11)*n=70
(24-2n)*n=70
2n^2-24n+70=0
n^2-12+35=0
(n-5)(n-7)=0
n=5或n=7

当n=5时
a1=13-2n=13-2*5=3

当n=7时
a1=13-2n=13-2*7=-1

Answer 5

Sn=n^2+5n-1
Sn-1=(n-1)^2+(n-1)-1
an=Sn-Sn-1=n^2+5n-1-[(n-1)^2+(n-1)-1]=2n+4
在等差数列{an}中，已知d=2,an=11 Sn=35，求an和n?求a1和n
sn=nan-1/2[n(n-1)d]
n^2-12n+35=0
n=5 n=7
a1=3 a1=-1