2018 AMC 12B Problems/Problem 9

Problem

What is $\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?$

$\textbf{(A) }100,100 \qquad \textbf{(B) }500,500\qquad \textbf{(C) }505,000 \qquad \textbf{(D) }1,001,000 \qquad \textbf{(E) }1,010,000 \qquad$

Solution 1

We can start by writing out the first couple of terms:

$(1+1) + (1+2) + (1+3) + \dots + (1+100)$ $(2+1) + (2+2) + (2+3) + \dots + (2+100)$ $(3+1) + (3+2) + (3+3) + \dots + (3+100)$ $\vdots$ $(100+1) + (100+2) + (100+3) + \dots + (100+100)$

Looking at the second terms in the parentheses, we can see that $1+2+3+\dots+100$ occurs $100$ times. It goes horizontally and exists $100$ times vertically. Looking at the first terms in the parentheses, we can see that $1+2+3+\dots+100$ occurs $100$ times. It goes vertically and exists $100$ times horizontally.

Thus, we have: $2\left(\dfrac{1+100}{2}\cdot 100\right).$

This gives us: $\boxed{\textbf{D} 1010000}.$

Solution 2

$\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} 100i+5050 = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{1,010,000}$ (Vfire)

Solution 3

$\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = (100)*(5050*2) = \boxed{1,010,000}$

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2018 AMC 12B Problems/Problem 9

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also