1958 AMC 12 第 26 题

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26.

一组 nn 个数的和为 ss。将每个数先加 2020,再乘 55,最后减 2020。所得新一组数的和为:

A set of nn numbers has the sum s.s. Each number of the set is increased by 20,20, then multiplied by 5,5, and then decreased by 20.20. The sum of the numbers in the new set thus obtained is:

s+20ns+20n

5s+80n5s+80n

ss

5s5s

5s+4n5s+4n

答案:B
知识点:分配律求和
难度评级:1390
小提示:

对原数中的一个一般项 uu 依次执行三步运算

Apply all three operations to a typical original number uu

大提示:

化简变换后的数,再对全部 nn 项求和

After simplifying the transformed number, sum over all nn entries

解答:

原数 uu 变为 5(u+20)20=5u+80 5(u+20)-20=5u+80\text{。}因而对原来的 nn 个数求和可得 5s+80n 5s+80n\text{。}

所以正确答案为 B

An original number uu becomes 5(u+20)20=5u+80. 5(u+20)-20=5u+80. Summing over the nn original numbers therefore gives 5s+80n. 5s+80n.

Thus, the correct answer is B.

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