(1)欲求1+3+32+33+…+320的值,可令S=1+3+32+33+…+320…①,将①式两边同乘以3,得______…②,由②式减去①式,得S=______.
(2)仿照(1)的方法,当k≠1时,试求a+ak+ak2+ak3+…akn的值(用含a,n,k的代数式表示)
人气:174 ℃ 时间:2019-08-17 12:13:12
解答
(1)3S=3+32+…+321…②,②-①得:2S=321-1,即S=12(321-1);(2)设M=a+ak+ak2+ak3+…akn,①两边都乘以k得:kM=ak+ak2+ak3+…akn+1,②②-①得:(k-1)M=akn+1-a,即M=a(kn+1−1)k−1,则a+ak+ak2+ak3+…akn=...
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