# Difference between revisions of "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: $$\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\dots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\dots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\dots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\dots &+ &(100+100) \end{array}$$ 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. 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.

Thus, we have $$2\left(\dfrac{100\cdot101}{2}\cdot 100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.$$

## Solution 2

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ &= \sum^{100}_{i=1} (100i+5050) \\ &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ &= 100\cdot5050+5050\cdot100 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~Vfire ~MRENTHUSIASM

## Solution 3

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that $$\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i = 100\cdot(5050\cdot2) = \boxed{\textbf{(E) }1{,}010{,}000}.$$

## Solution 4

The minimum term is $1 + 1 = 2$, and the maximum term is $100 + 100 = 200$. The average of the $100 \cdot 100 = 10{,}000$ terms is the average of the minimum and maximum terms, which is $\frac{2+200}{2}=101$. The sum is therefore $101 \cdot 10{,}000 = \boxed{\textbf{(E) }1{,}010{,}000}$.