Difference between revisions of "2018 AMC 12B Problems/Problem 9"
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==Solution 7== | ==Solution 7== | ||
− | Quick and Easy: Note that | + | Quick and Easy: Note that the full summation can be evaluated from the inside out. First 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> We begin with the inside: |
<cmath>\begin{align*} \sum^{100}_{j=1} (i+j)=100i+5050 \end{align*}</cmath> | <cmath>\begin{align*} \sum^{100}_{j=1} (i+j)=100i+5050 \end{align*}</cmath> | ||
Revision as of 08:43, 4 February 2022
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
Solution 7
Quick and Easy: Note that the full summation can be evaluated from the inside out. First recall that the sum of the first positive integers is We begin with the inside:
Then take that result and evaluate the outer sigma:
~Directrixxx
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.