Difference between revisions of "2018 AMC 12B Problems/Problem 9"
MRENTHUSIASM (talk | contribs) (Improved Sol 3 significantly.) |
MRENTHUSIASM (talk | contribs) (→Solution 4: Rewrote the claim of average to be clearer and more concise.) |
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== Solution 4 == | == Solution 4 == | ||
− | The | + | 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> |
+ | ~Rejas ~MRENTHUSIASM | ||
==See Also== | ==See Also== |
Revision as of 13:33, 20 September 2021
Problem
What is
Solution 1
We can start by writing out the first couple of terms: Looking at the first terms in the parentheses, we can see that occurs times. It goes vertically and exists times horizontally. Looking at the second terms in the parentheses, we can see that occurs times. It goes horizontally and exists times vertically.
Thus, we have
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 ~RandomPieKevin ~MRENTHUSIASM
Solution 4
The sum contains terms, and the average value of both and is Therefore, the sum becomes ~Rejas ~MRENTHUSIASM
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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