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# 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 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 $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ Since the nested summation is symmetric with respect to $i$ and $j,$ it follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\ &= 2\sum^{100}_{i=1} \sum^{100}_{i=1} i \\ &= 2\sum^{100}_{i=1} 5050 \\ &= 2\cdot(5050\cdot100) \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~Vfire ~MRENTHUSIASM

## Solution 4

The sum contains $100\cdot100=10000$ terms, and the average value of both $i$ and $j$ is $\frac{101}{2}.$ Therefore, the sum becomes $$10000\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.$$ ~Rejas ~MRENTHUSIASM

## Solution 5

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 6

When we expand the nested summation as shown in Solution 5, note that:

1. The term $2$ appears $1$ time.

The term $3$ appears $2$ times.

The term $4$ appears $3$ times.

$\cdots$

The term $101$ appears $100$ times.

More generally, the term $k+1$ appears $k$ times for $k\in\{1,2,3,\ldots,100\}.$

2. The term $102$ appears $99$ times.

The term $103$ appears $98$ times.

The term $104$ appears $97$ times.

$\cdots$

The term $200$ appears $1$ time.

More generally, the term $k+101$ appears $100-k$ times for $k\in\{1,2,3,\ldots,99\}.$

Together, the nested summation becomes \begin{align*} \sum^{100}_{k=1}[(k+1)k] + \sum^{99}_{k=1}[(k+101)(100-k)] &= \sum^{100}_{k=1}[k^2+k] + \sum^{99}_{k=1}[-k^2-k+10100] \\ &= \sum^{100}_{k=1}k^2 + \sum^{100}_{k=1}k - \sum^{99}_{k=1}k^2 - \sum^{99}_{k=1}k + \sum^{99}_{k=1}10100 \\ &= \left(\sum^{100}_{k=1}k^2 - \sum^{99}_{k=1}k^2\right) + \left(\sum^{100}_{k=1}k - \sum^{99}_{k=1}k\right) + \sum^{99}_{k=1}10100 \\ &= 100^2+100+999900 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~MRENTHUSIASM

## See Also

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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