Difference between revisions of "1992 AHSME Problems/Problem 21"
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\text{(E) } 1009</math> | \text{(E) } 1009</math> | ||
− | == Solution == | + | == Solution 1== |
Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is | Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is | ||
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<cmath>= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)</cmath> | <cmath>= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)</cmath> | ||
<cmath>= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.</cmath> | <cmath>= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.</cmath> | ||
− | If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is <math>1000 * 99 = 99,000,</math> than the Cesáro total of the second sequence is <math>n \cdot a_1 + 99,000 = 100 | + | If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is <math>1000 * 99 = 99,000,</math> than the Cesáro total of the second sequence is <math>n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.</math> Thus the Cesáro sum of the second sequence is <math>\frac{99,100}{100} = \boxed{991, A}\, .</math> |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We know that the Cesáro sum of the first <math>99</math> numbers is | ||
+ | |||
+ | <cmath>\frac{a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}}{99}=100.</cmath> | ||
+ | |||
+ | Multiplying by <math>99</math>, we see that | ||
+ | |||
+ | <cmath>a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}=100\cdot{99}.</cmath> | ||
+ | |||
+ | Now, we append <math>1</math> to the list of <math>a_n,</math> so our new sum becomes | ||
+ | |||
+ | <cmath>\frac{1+1+a_1+1+a_1+a_2+...+a_{99}}{100} \rightarrow</cmath> | ||
+ | <cmath>\frac{100+a_1+a_1+a_2+...+a_{99}}{100}.</cmath> | ||
+ | |||
+ | We've estabished that <math>a_1+a_1+a_2+a_1+a_2+a_3+...+a_{99}=100\cdot{99}</math>, so | ||
+ | |||
+ | <cmath>\frac{100+a_1+a_1+a_2+...+a_{99}}{100}=\frac{100+99\cdot{100}}{100}.</cmath> | ||
+ | |||
+ | Dividing everything by <math>100</math>, the sum of the sought sequence is <math>1+990=991 \rightarrow \boxed{A}</math> | ||
+ | |||
+ | ~Benedict T (countmath1) | ||
== See also == | == See also == |
Latest revision as of 21:53, 17 April 2022
Contents
Problem
For a finite sequence of numbers, the Cesáro sum of A is defined to be , where and . If the Cesáro sum of the 99-term sequence is 1000, what is the Cesáro sum of the 100-term sequence ?
Solution 1
Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is than the Cesáro total of the second sequence is Thus the Cesáro sum of the second sequence is
Solution 2
We know that the Cesáro sum of the first numbers is
Multiplying by , we see that
Now, we append to the list of so our new sum becomes
We've estabished that , so
Dividing everything by , the sum of the sought sequence is
~Benedict T (countmath1)
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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