已知数列an,an属于N+,Sn＝(1/8)*(an +2)^2
1个回答

8Sn＝(an +2)^2

8Sn-1＝(an-1 +2)^2

而an= sn - sn-1

8an = (an +2)^2 - (an-1 +2)^2

0 = (an -2)^2 - (an-1 +2)^2

=>

(an -2)^2 = (an-1 +2)^2

=> an - 2 = ±(an-1 +2)

若 an - 2 = -(an-1 +2)

则 an = -an-1（正负交替或一直为0）

与an属于N+矛盾

故an - 2 = +(an-1 +2)

=>an = an-1 +4

即an是公差为4的等差数列

由于S1=a1,故

8a1=a1²+4a1+4

0=(a1-2)²

a1=2

=>

b1=-29

所以bn是公差为2,首项为-29的等差数列

显然加到-1是最小（再往后算就是加正数了,当然会变大）

故最小值=-1~-29

= -30*15/2

= -225

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