Difference between revisions of "2018 AMC 12B Problems/Problem 9"
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== Solution 5 == | == Solution 5 == | ||
− | We | + | We start by writing out the first few terms: |
<cmath>\begin{array}{ccccccccc} | <cmath>\begin{array}{ccccccccc} | ||
(1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ | (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ | ||
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(3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] | (3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] | ||
&&&&\vdots&&&& \\ | &&&&\vdots&&&& \\ | ||
− | (100+1) &+ &(100+2) &+ &(100+3) &+ &\ | + | (100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100). |
\end{array}</cmath> | \end{array}</cmath> | ||
− | From the first terms in the parentheses, the sum <math>1+2+3+\ | + | From the first terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times vertically. |
− | From the second terms in the parentheses, the sum <math>1+2+3+\ | + | From the second terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times horizontally. |
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> | 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> | ||
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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 \\ | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/flvmGmDmwZ8 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== |
Latest revision as of 22:03, 27 May 2023
Contents
Problem
What is
Solution 1
Recall that the sum of the first positive integers is It follows that ~MRENTHUSIASM
Solution 2
Recall that the sum of the first positive integers is It follows that ~Vfire ~MRENTHUSIASM
Solution 3
Recall that the sum of the first positive integers is Since the nested summation is symmetric with respect to and it follows that ~Vfire ~MRENTHUSIASM
Solution 4
The sum contains terms, and the average value of both and is Therefore, the sum becomes ~Rejas ~MRENTHUSIASM
Solution 5
We start by writing out the first few terms: From the first terms in the parentheses, the sum occurs times vertically.
From the second terms in the parentheses, the sum occurs times horizontally.
Recall that the sum of the first positive integers is Therefore, the answer is ~RandomPieKevin ~MRENTHUSIASM
Solution 6
When we expand the nested summation as shown in Solution 5, note that:
- The term occurs time.
The term occurs times.
The term occurs times.
The term occurs times.
More generally, the term occurs times for
- The term occurs times.
The term occurs times.
The term occurs times.
The term occurs time.
More generally, the term occurs times for
Together, the nested summation becomes ~MRENTHUSIASM
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
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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