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数学
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设数列5,12,20.,前N项和Sn=a^2+bn+c(1)求a,b,c的值
人气:301 ℃ 时间:2020-01-28 20:15:08
解答
Sn=an^2+bn+c
n=1,S1=a+b+c=a1=5 (1)
n=2,S2=4a+2b+c=a1+a2=17 (2)
n=3,S3=9a+3b+c=a1+a2+a3=37 (3)
(2)-(1):3a+b=12
(3)-(2):5a+b=20
解得:a=4,b=0,c=1
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