AMC12 2020 B

AMC12 2020 B · Q1

AMC12 2020 B · Q1. It mainly tests Fractions, Perfect squares & cubes.

What is the value in simplest form of the following expression? $\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$

下列表达式的值的最简形式是多少？ $\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$

(A) 5 5

(B) $4 + \sqrt{7} + \sqrt{10}$ $4 + \sqrt{7} + \sqrt{10}$

(D) 15 15

(E) $4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$ $4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

A direct calculation shows $\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16} = 1 + 2 + 3 + 4 = 10$. Note: This problem illustrates the fact that the sum of the first $n$ odd integers equals $n^2$.

直接计算显示 $\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16} = 1 + 2 + 3 + 4 = 10$。注：此题说明前$n$个奇数的和等于$n^2$。

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