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Difference between revisions of "2018 AMC 12B Problems/Problem 9"

(→‎Solution 4: Rewrote the claim of average to be clearer and more concise.)
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The sum contains <math>100\cdot100=10000</math> terms, and the average value of both <math>i</math> and <math>j</math> is <math>\frac{101}{2}.</math> Therefore, the sum becomes <cmath>10000\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
 
The sum contains <math>100\cdot100=10000</math> terms, and the average value of both <math>i</math> and <math>j</math> is <math>\frac{101}{2}.</math> Therefore, the sum becomes <cmath>10000\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
 
~Rejas ~MRENTHUSIASM
 
~Rejas ~MRENTHUSIASM
 +
 +
== Solution 5 ==
 +
When we expand the nested summation as shown in Solution 1, note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>The term <math>2</math> appears <math>1</math> time. <p>
 +
The term <math>3</math> appears <math>2</math> times. <p>
 +
The term <math>4</math> appears <math>3</math> times. <p>
 +
<math>\cdots</math> <p>
 +
The term <math>101</math> appears <math>100</math> times. <p>
 +
More generally, the term <math>k+1</math> appears <math>k</math> times for <math>k\in\{1,2,3,\ldots,100\}.</math><p></li>
 +
  <li>The term <math>102</math> appears <math>99</math> times. <p>
 +
The term <math>103</math> appears <math>98</math> times. <p>
 +
The term <math>104</math> appears <math>97</math> times. <p>
 +
<math>\cdots</math> <p>
 +
The term <math>200</math> appears <math>1</math> time. <p>
 +
More generally, the term <math>k+101</math> appears <math>100-k</math> times for <math>k\in\{1,2,3,\ldots,99\}.</math><p></li>
 +
</ol>
 +
Together, the nested summation becomes
 +
<cmath>\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*}</cmath>
 +
~MRENTHUSIASM
  
 
==See Also==
 
==See Also==

Revision as of 14:28, 20 September 2021

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 $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ 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*} ~RandomPieKevin ‎~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

When we expand the nested summation as shown in Solution 1, 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 by
Problem 8 		Followed by
Problem 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

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