Question

设数列｛an}的前n项和为Sn，Sn={a1[(3^n)-1]}/2 (对于所有n≥1)，且a5=54,则a1的数值是？

Answer 1

a5=S5-S4=[(3^5-1)/2]*a1-(3^4-1)/悔孙并碧迹2]*a1=81a1=54
就是把n=4和5代入Sn的式子里级OK了凯握，
所以解得a1=2/3

Answer 2

a1=2/键档3
由题目可知
S1=a1=a1(3^1-1)/2
S2=a1(3^2-1)/2=4*a1=a1+3*a1
S3=a1(3^3-1)/稿卖2=13*a1=a1+3*a1+9*a1
S4=a1(3^4-1)/2=40*a1=a1+3*a1+9*a1+27*a1
S5=a1(3^5-1)/2=121*a1=a1+3*a1+9*a1+27*a1+81*a1
S5-S4=a5
所以a5=81*a1=54
那么稿敬乱a1=2/3