AMC12 2018 B

AMC12 2018 B · Q9

AMC12 2018 B · Q9. It mainly tests Basic counting (rules of product/sum).

What is \sum_{i=1}^{100} \sum_{j=1}^{100} (i + j)?

\sum_{i=1}^{100} \sum_{j=1}^{100} (i + j) 的值为多少？

(A) 100,100 100,100

(B) 500,500 500,500

(D) 1,001,000 1,001,000

(E) 1,010,000 1,010,000

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Note that the sum of the first 100 positive integers is $\frac{1}{2}\cdot 100 \cdot 101 = 5050$. Then $$ \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) = \sum_{i=1}^{100}\sum_{j=1}^{100} i + \sum_{i=1}^{100}\sum_{j=1}^{100} j $$ $$ = \sum_{j=1}^{100}\sum_{i=1}^{100} i + \sum_{i=1}^{100}\sum_{j=1}^{100} j $$ $$ = 100\sum_{i=1}^{100} i + 100\sum_{j=1}^{100} j $$ $$ = 100(5050+5050) = 1{,}010{,}000. $$

答案（E）：注意，前 $100$ 个正整数的和为 $\frac{1}{2}\cdot 100 \cdot 101 = 5050$。那么 $$ \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) = \sum_{i=1}^{100}\sum_{j=1}^{100} i + \sum_{i=1}^{100}\sum_{j=1}^{100} j $$ $$ = \sum_{j=1}^{100}\sum_{i=1}^{100} i + \sum_{i=1}^{100}\sum_{j=1}^{100} j $$ $$ = 100\sum_{i=1}^{100} i + 100\sum_{j=1}^{100} j $$ $$ = 100(5050+5050) = 1{,}010{,}000. $$

Topics

CP.001 Basic counting (rules of product/sum)