Question

高一数学

已知数列an满足a1等于1an>0Sn是数列的前n项和对任意N属于N*有2sn=2an方+an-1
1计算a2，a3的值并求数列an的通项公式
2求满足Sn小于等于27的m最大值

Answer 1

(1) 2Sn=2an^2+an-1
2Sn+1=2an+1 ^2+ an+1 -1
两式相减，得
2(Sn+1 - Sn)=2(an+1 ^2 -an ^2)+(an+1 - an)
即，2an+1=2(an+1 ^2 -an ^2)+(an+1 - an)
an+1 + an=2(an+1 + an)(an+1 - an)
an+1 - an =1/2
所以，{an}是公差为1/2的等差数列
an=a1+(n-1)*1/2=1+(n-1)/2=(n+1)/2
a2=(2+1)/2=3/2， a3=(3+1)/2=2

(2) Sn=n*(a1+an)/2=n*(1+(n+1)/2)/2=n(n+3)/4
Sn<=27
即，n(n+3)/4<=27
n^2+3n-108<=0
(n-9)(n+12)<=0
-12<=n<=9
所以，满足Sn<=27的n的最大值为9

Answer 2