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

Latest revision as of 00:45, 5 February 2022

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:

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 $k\in\{1,2,3,\ldots,100\}.$
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

@@ Line 49: / Line 49: @@
 == Solution 5 ==
-We can start by writing out the first few terms:
+We start by writing out the first few terms:
 <cmath>\begin{array}{ccccccccc}
 (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\

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

Page

Toolbox

Search