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2018 AMC 12B Problems/Problem 9: Difference between revisions

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<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath>
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath>


<math> \textbf{(A) }100,100 \qquad
<math> \textbf{(A) }100{,}100 \qquad
\textbf{(B) }500,500\qquad
\textbf{(B) }500{,}500\qquad
\textbf{(C) }505,000 \qquad
\textbf{(C) }505{,}000 \qquad
\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 ==
== 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


We can start by writing out the first couple of terms:
== Solution 2 ==
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} \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*}</cmath>
~Vfire ~MRENTHUSIASM


<cmath>(1+1) + (1+2) + (1+3) + \dots + (1+100)</cmath>
== Solution 3 ==
<cmath>(2+1) + (2+2) + (2+3) + \dots + (2+100)</cmath>
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> Since the nested summation is symmetric with respect to <math>i</math> and <math>j,</math> it follows that
<cmath>(3+1) + (3+2) + (3+3) + \dots + (3+100)</cmath>
<cmath>\begin{align*}
<cmath>\vdots</cmath>
\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\
<cmath>(100+1) + (100+2) + (100+3) + \dots + (100+100)</cmath>
&= 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*}</cmath>
~Vfire ~MRENTHUSIASM


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.
== Solution 4 ==
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.
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


Thus, we have: <cmath>2\left(\dfrac{1+100}{2}\cdot 100\right).</cmath>
== Solution 5 ==


This gives us: <cmath>\boxed{\textbf{D} 1010000}.</cmath>
We start by writing out the first few terms:
<cmath>\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}</cmath>
From the first terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times vertically.


== Solution 2 ==
From the second terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times horizontally.


<cmath> \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} </cmath> (Vfire)
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 ==
When we expand the nested summation as shown in Solution 5, note that:
<ol style="margin-left: 1.5em;">
  <li>The term <math>2</math> occurs <math>1</math> time. <p>
The term <math>3</math> occurs <math>2</math> times. <p>
The term <math>4</math> occurs <math>3</math> times. <p>
<math>\cdots</math> <p>
The term <math>101</math> occurs <math>100</math> times. <p>
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> occurs <math>99</math> times. <p>
The term <math>103</math> occurs <math>98</math> times. <p>
The term <math>104</math> occurs <math>97</math> times. <p>
<math>\cdots</math> <p>
The term <math>200</math> occurs <math>1</math> time. <p>
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>
Together, the nested summation becomes
<cmath>\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*}</cmath>
~MRENTHUSIASM


== Solution 3 ==
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 
https://youtu.be/flvmGmDmwZ8
<cmath> \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}  </cmath>


~Education, the Study of Everything


==See Also==
==See Also==

Latest revision as of 21:03, 27 May 2023

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

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/flvmGmDmwZ8

~Education, the Study of Everything

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
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