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

(Rearranged the solutions based on similar ideas.)
m (Solution 5)
 
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\textbf{(D) }1{,}001{,}000 \qquad
 
\textbf{(D) }1{,}001{,}000 \qquad
 
\textbf{(E) }1{,}010{,}000 \qquad </math>
 
\textbf{(E) }1{,}010{,}000 \qquad </math>
 +
 +
== Solution 1 ==
 +
Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that
 +
<cmath>\begin{align*}
 +
\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\
 +
&= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\
 +
&= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\
 +
&= 100\cdot5050 + 100\cdot5050 \\
 +
&= \boxed{\textbf{(E) }1{,}010{,}000}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM
  
 
== Solution 2 ==
 
== Solution 2 ==
Line 33: Line 44:
  
 
== Solution 4 ==
 
== Solution 4 ==
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\cdot\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
 
~Rejas ~MRENTHUSIASM
 
~Rejas ~MRENTHUSIASM
  
 
== Solution 5 ==
 
== Solution 5 ==
  
We can start by writing out the first couple of terms:
+
We start by writing out the first few terms:
 
<cmath>\begin{array}{ccccccccc}
 
<cmath>\begin{array}{ccccccccc}
(1+1) &+ &(1+2) &+ &(1+3) &+ &\dots &+ &(1+100) \\
+
(1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\
(2+1) &+ &(2+2) &+ &(2+3) &+ &\dots &+ &(2+100) \\
+
(2+1) &+ &(2+2) &+ &(2+3) &+ &\cdots &+ &(2+100) \\
(3+1) &+ &(3+2) &+ &(3+3) &+ &\dots &+ &(3+100) \\ [-1ex]
+
(3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex]
 
&&&&\vdots&&&& \\
 
&&&&\vdots&&&& \\
(100+1) &+ &(100+2) &+ &(100+3) &+ &\dots &+ &(100+100)
+
(100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100).
 
\end{array}</cmath>
 
\end{array}</cmath>
Looking at the first terms in the parentheses, we can see that <math>1+2+3+\dots+100</math> occurs <math>100</math> times. It goes vertically and exists <math>100</math> times horizontally.
+
From the first terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times vertically.
Looking at the second terms in the parentheses, we can see that <math>1+2+3+\dots+100</math> occurs <math>100</math> times. It goes horizontally and exists <math>100</math> times vertically.
+
 
 +
From the second terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times horizontally.
  
Thus, we have <cmath>2\left(\dfrac{100\cdot101}{2}\cdot 100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
+
Recall that the sum of the first <math>100</math> positive integers is <math>1+2+3+\cdots+100=\frac{101\cdot100}{2}=5050.</math> Therefore, the answer is <cmath>2\cdot\left(5050\cdot100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
 +
~RandomPieKevin ‎~MRENTHUSIASM
  
 
== Solution 6 ==
 
== Solution 6 ==
 
When we expand the nested summation as shown in Solution 5, note that:
 
When we expand the nested summation as shown in Solution 5, note that:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
   <li>The term <math>2</math> appears <math>1</math> time. <p>
+
   <li>The term <math>2</math> occurs <math>1</math> time. <p>
The term <math>3</math> appears <math>2</math> times. <p>
+
The term <math>3</math> occurs <math>2</math> times. <p>
The term <math>4</math> appears <math>3</math> times. <p>
+
The term <math>4</math> occurs <math>3</math> times. <p>
 
<math>\cdots</math> <p>
 
<math>\cdots</math> <p>
The term <math>101</math> appears <math>100</math> times. <p>
+
The term <math>101</math> occurs <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>
+
More generally, the term <math>k+1</math> occurs <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>
+
   <li>The term <math>102</math> occurs <math>99</math> times. <p>
The term <math>103</math> appears <math>98</math> times. <p>
+
The term <math>103</math> occurs <math>98</math> times. <p>
The term <math>104</math> appears <math>97</math> times. <p>
+
The term <math>104</math> occurs <math>97</math> times. <p>
 
<math>\cdots</math> <p>
 
<math>\cdots</math> <p>
The term <math>200</math> appears <math>1</math> time. <p>
+
The term <math>200</math> occurs <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>
+
More generally, the term <math>k+101</math> occurs <math>100-k</math> times for <math>k\in\{1,2,3,\ldots,99\}.</math><p></li>
 
</ol>
 
</ol>
 
Together, the nested summation becomes  
 
Together, the nested summation becomes  
 
<cmath>\begin{align*}
 
<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}\left[(k+1)k\right] + \sum^{99}_{k=1}\left[(k+101)(100-k)\right] &= \sum^{100}_{k=1}\left[k^2+k\right] + \sum^{99}_{k=1}\left[-k^2-k+10100\right] \\
 
&= \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 \\
 
&= \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 \\
 
&= \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 \\
+
&= 100^2+100+10100\cdot99 \\
 
&= \boxed{\textbf{(E) }1{,}010{,}000}.
 
&= \boxed{\textbf{(E) }1{,}010{,}000}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Latest revision as of 12:26, 24 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

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} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ &= 100\cdot5050 + 100\cdot5050 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~MRENTHUSIASM

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 \[10000\cdot\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.\] ~Rejas ~MRENTHUSIASM

Solution 5

We start by writing out the first few terms: \[\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\cdots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100). \end{array}\] From the first terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times vertically.

From the second terms in the parentheses, the sum $1+2+3+\cdots+100$ occurs $100$ times horizontally.

Recall that the sum of the first $100$ positive integers is $1+2+3+\cdots+100=\frac{101\cdot100}{2}=5050.$ Therefore, the answer is \[2\cdot\left(5050\cdot100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.\] ~RandomPieKevin ‎~MRENTHUSIASM

Solution 6

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

  1. The term $2$ occurs $1$ time.

    The term $3$ occurs $2$ times.

    The term $4$ occurs $3$ times.

    $\cdots$

    The term $101$ occurs $100$ times.

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

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

    The term $103$ occurs $98$ times.

    The term $104$ occurs $97$ times.

    $\cdots$

    The term $200$ occurs $1$ time.

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

Together, the nested summation becomes \begin{align*} \sum^{100}_{k=1}\left[(k+1)k\right] + \sum^{99}_{k=1}\left[(k+101)(100-k)\right] &= \sum^{100}_{k=1}\left[k^2+k\right] + \sum^{99}_{k=1}\left[-k^2-k+10100\right] \\ &= \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+10100\cdot99 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~MRENTHUSIASM

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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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